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HARVARD UNIVERSITY 

Library of the 

Museum of 

Comparative Zoology 



THE UNIVERSITY OF KANSAS 
MUSEUM OF NATURAL HISTORY 



SPECIAL PUBUCATION 
No, 19 



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A Primer of Phylogenetic Procedmre^ 



E. O. Wiley 

D. Sdegel-Causey 

D. R. Brooks 

V. A. Funk 



UWA/RENCE 



October 1991 






The University of Kansas 

Museum of Natural History 

Special Publications 



Museim of Comparative Zoology Library 

Harvard Univsr^ity 



To receive our 1990 Catalog of Publications, contact Publications, 
Museum of Natural History, The University of Kansas, Lawrence, 
Kansas 66045-2454, USA. To order by phone call 913-864-4540. 
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other available numbers in this series. 



The University of Kansas 
Museum of Natural History 

Special Publication No. 19 
October 1991 



The Compleat Cladkt 

A Primer of Phylogenetic Procedures 



E. O. Wiley 

Museum of Natural Histon/ 

The University ofKnmas 

Lawrence, Kansas 66045 



D. Siegel-Causey 

Museum of Natural Histonj 
Hie University of Kansas 
Lawrence, Kansas 66045 



D. R. Brooks 

Department of Zoology 

Tlie Universit]/ of Toronto 

Toronto, Ontario M5S lAl 

CANADA 



V. A. Funk 

Department of Botany 

National Museum of Natural Histori/ 

Tlie Smitliso)iian Institution 

Washington, D.C. 20560 



Museum of Natural Hjstory 

Dyche Hall 

The University of Kansas 

Lawrence, Kansas 

1991 



n 



-^ The University of Kansas 

\^^ I Museum of Natural History 



6,6 A^^^ 

Joseph T. Collins, Editor 
iAtP^^'^^vVS Kimberlee Wollter. Copyediting and Design 
yA^jjiVC*^ Kate Shaw, Design and Typesetting 



© 1991 Museum of Natural History 

The University of Kansas 

Lawrence, Kansas 66045-2454, USA 



Special Publication No. 19 

pp. X + 1-158; 118 figures; 58 tables 

Published October 1991 

ISBN 0-89338-035-0 



Text for this publication was produced on a Macintosh 11 computer in Microsoft® 
Word, and figures were drafted in Claris^"' MacDraw- II. The publication was then 
designed and typeset in Aldus PageMaker® and forwarded to the printer. 



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ERRATA 

The Compleat Cladist: A Primer ofPhylogenetic Procedures 

by 
E. O. Wiley, D. Siegel-Causey, D. R. Brooks, and V. A. Funk 



On p. 74 the formula for "r" should read: 

r = ^^^ 
r-g-m 



On p. 107, change legend for Figure 6.17 to read "A 

phylogeny of hypothetical species of the genus Mus, . 



The University of Kansas 
Museum of Natural History 
Dyche Hall 
Lawrence, KS 66045-2454 



Nonprofit Org. 
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Lawrence, KS 
Permit No. 65 



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"... but, he that hopes to be a good Angler must not onely bring an 
inquiring, searching, observing wit, but he must bring a large measure 
of hope and patience, and a love and propensity to the Art it self; but 
having once got and practis'd it, then doubt not but Angling will prove 
to be so pleasant, that it will prove like Vertue, a reward to it self." 



Piscator speaking to Venator and Auceps 

The Compleat Angler by Izaak Walton 

The Modem Library Printing of the Fourth (1668) Edition 

Random House, New York 



ui 



IV 



PREFACE 



In writing this worlcbook, we have strived to follow in the tradition of Brooks et al. (1984) 
of providing a guide to basic phylogenetic techniques as we now understand them. The field 
of phylogenetics has undergone many changes, some philosophical and some empirical, in 
the last 10 years. We hope to reflect some of these changes in this workbook. 

The workbook is arranged in a manner roughly functional and pedagogical. The 
sophisticated reader, for example, might question why we spend so much time with 
exercises (in Chapter 2) that do not really reflect the way "real" phylogenetic analysis is 
performed or why so much space is given to Hennig Argumentation and Wagner algorithms 
(Chapter 4) when they are either not a part of modem computer algorithms or, at best, are 
only a starting point for finding the best hypothesis of common ancestry for the particular 
data analyzed. Our answer, perhaps constrained by our own histories, is that this approach 
seems to help at least some students learn phylogenetics. 

Because we can only touch on the most basic topics and provide exercises for only a few 
of these, we invite the student to explore the original literature; we have cited sources in the 
body of the text and called attention to some papers in the "Chapter Notes and References" 
sections at the end of each chapter. The absence of a paper in the text or in the "Notes" 
section is no reflection on the worth of the paper. A compilation of all of the useful papers 
relating to phylogenetics is quite beyond the scope of this workbook. 

In addidon to the exercises, we have provided immediate feedback sections termed 
"Quick Quizzes." We are interested in reader opinion regarding both the exercises and the 
quick quizzes. We will incorporate suggestions, wherever possible, in subsequent editions. 

The major title. The Compleat Cladist. is inspired by the title The Compleat Angler by 
Izaak Walton, a marvelous book published in many editions since 1653. Of course, this book 
is not "complete" or even "compleat" in the archaic sense of representing a book that teaches 
complete mastery of a subject. Phylogenetics is much too dynamic for a small workbook to 
fulfill that criterion. Rather, we take our inspiration from Walton; the compleat cladist is one 
who approaches the subject with energy, wonder, and joy. Unfortunately, none of us are 
clever enough to come up with an analogy to the "Anglers Song." 

We thank the following people for their valuable comments on part or all of the earlier 
drafts of this workbook: the students of Biology 864 and Mike Bamshad (University of 
Kansas), John Hayden (University of Richmond), Debbie McLennan (University of 
Toronto), David Swofford (Illinois Natural History Survey), Charlotte Taylor, Richard 
Thomas, and Rafael Joglar (University of Puerto Rico), Wayne Maddison (Harvard Univer- 
sity), and Arnold Kluge (University of Michigan). 

Special thanks are due to David Kizirian (University of Kansas) for working through the 
answers to the exercises and to Kate Shaw and Kim Wollter (University of Kansas) for their 



editorial skills. Partial support in the form of computer hardware and software by the 
National Science Foundation (BSR 8722562) and the University of Kansas Museum of 
Natural History is gratefully acknowledged. Mistakes in interpretation and exercise answers 
are our responsibility, and we will be grateful for any suggestions and corrections for 
incorporation into future editions. 



E. O. Wiley, D. Siegel-Causey, D. R. Brooks, and V. A. Funk 
Lawrence, Kansas; Toronto, Ontario; and Washington, D.C. 
Suinmer, 1991 



VI 



CONTENTS 



PREFACE V 

CHAPTER 1 : INTRODUCTION, TERMS, AND CONCEPTS 1 

Terms for Groups of Organisms 3 

Quick Quiz — Groups 5 

Terms for the Relationships of Taxa 6 

Quick Quiz — Relationships 7 

Terms for Classifications 7 

Quick Quiz — Classification 8 

Process Terms 8 

Terms for the Attributes of Specimens 9 

Quick Quiz — Characters 11 

Chapter Notes and References 11 

Quick Quiz Answers 11 

CHAPTER 2: BASIC PHYLOGENETIC TECHNIQUES 13 

Quick Quiz — Basic Rules of Analysis 17 

Sample Analyses 17 

Exercises 22 

Chapter Notes and References 24 

Quick Quiz Answers 24 

CHAPTER 3: CHARACTER ARGUMENTATION AND CODING 25 

Outgroup Comparison 25 

Polarity Decisions 25 

Rules of Thumb 31 

Other Situations 31 

Quick Quiz — Outgroups and Polarities 31 

Polarity Exercises 32 

Character Coding 34 

Quick Quiz — Character Coding 40 

Coding Exercises 40 

Chapter Notes and References 42 

Quick Quiz Answers 43 

CHAPTER 4: TREE BUILDING AND OPTIMIZATION 45 

Hennig Argumentation 45 

Hennig Exercises 46 



vu 



The Wagner Algorithm 47 

Wagner Definitions 49 

The Algorithm 50 

Wagner Tree Exercises 54 

Optimal Trees and Parsimony Criteria 54 

Optimizing Trees 56 

ACCTRAN 57 

ACCTRAN Exercises 60 

Discussion 61 

Finding MPR Sets 62 

DELTRAN 63 

DELTRAN Exercises 63 

Current Technology 66 

Chapter Notes and References 68 

CHAPTER 5: TREE COMPARISONS 71 

Summary Tree Measures 71 

Tree Length 71 

Consistency Indices 72 

Ensemble Consistency Indices 75 

TheF-Ratio 76 

Tree Summaries Exercises 78 

Consensus Techniques 80 

Strict Consensus Trees 81 

Adams Trees 83 

Majority Consensus Trees 88 

Chapter Notes and References 89 

CHAPTER 6: CLASSIFICATION 91 

Evaluation of Existing Classifications 92 

Logical Consistency 92 

Determining the Number of Derivative Classifications 99 

Classification Evaluation Exercises 100 

Constructing Phylogenetic Classifications 102 

Rules of Phylogenetic Classifications 102 

Conventions 103 

Quick Quiz — Taxonomy vs. Systematics 108 

Convention Exercises 108 

Chapter Notes and References Ill 

Quick Quiz Answers Ill 



CHAPTER 7: COEVOLUTIONARY STUDIES 113 

Coding Phylogenetic Trees 113 

Quick Quiz — ^Biogeography 115 

Single Tree Exercises 117 

More Than One Group 118 

Missing Taxa 120 

Widespread Species 124 

Sympatry within a Clade 127 

Tlie Analogy between Phylogenetics and Historical Biogeography 127 

Chapter Notes and References 128 

Quick Quiz Answers 128 

LITERATURE CITED 129 

ANSWERS TO EXERCISES 137 

Chapter 2 137 

Chapter 3 139 

Chapter 4 141 

Chapters 150 

Chapter 6 151 

Chapter 7 156 



IX 



Chapter 1 

INTRODUCTION, TERMS, AND CONCEPTS 



The core concept of phylogenetic systematics is the use of derived or apomorphic 
characters to reconstruct common ancestry relationships and the grouping of taxa based on 
common ancestry. This concept, first formahzed by Hennig ( 1950, 1966), has been slowly, 
and not so quietly, changing the nature of systematics. Why should we be interested in this 
approach? What about phylogenetic systematics is different from traditional systematics? 
The answer is simple: classifications that are not known to be phylogenetic are possibly 
artificial and are. therefore, useful only for identification and not for asking questions about 
evolution. 

There are two other means of making statements of relationship: traditional systematics 
and phenetics. Traditional systematic methods employ intuition. In practical terms, intuition 
is character weighting. The scientist studies a group of organisms, selects the character(s) 
believed to be important (i.e., conservative), and delimits species and groups of species 
based on these characters. Disagreements usually arise when different scientists think 
different characters are important. It is difficult to evaluate the evolutionary significance of 
groups classified by intuition because we do not know why they were created or whether 
they represent anything real in nature. Because these groups may not be defined at all or may 
be defined by characters that have no evolutionary significance, such groups may be 
artificial. 

Phenetics is an attempt to devise an empirical method for determining taxonomic 
relationships. In practice, phenetics is no better than traditional systemafics in determining 
relationships because the various algorithms concentrate on reflecting the total similarity of 
the organisms in question. Organisms that appear to be more similar are grouped together, 
ignoring the results of parallel or convergent evolution and again creating possibly artificial 
groups. 

Phylogeneticists differ from traditional systematists in that we employ empirical methods 
to reconstruct phylogenies and strictly evolutionary principles to form classifications rather 
than relying on intuition or authority. We differ from pheneticists in that our methods seek to 
find the genealogic relationships among the taxa we study rather than the phenetic or overall 
similarity relationships. 

What all this means is that the groups we discover are thought to be natural, or 
monophyletic. Given any array of taxa, which two are more closely related to each other 
than either is to any other taxon? We attempt to discover the common ancestry relationships 
indirectly through finding evidence for common ancestry. Tliis evidence comes in the fonn 
of shared derived characters (synapomorphies). For example, among Aves (birds), 
Crocodylia (alligators and crocodiles), and Squamata (lizards, snakes, and amphisbaenians), 
Aves and Crocodylia are thought to be more closely related because they share a number of 
synapomorphies thought to have originated in their common ancestor, which appeared after 



2 KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 

(later than) the common ancestor of all three taxa. This relationship is shown in the form of 
a phylogenetic tree, a reconstruction of the genealogic relationships. 

In addition, phylogeneticists view the reconstructed tree (frequently termed a cladogram) 
as the classification, and when expressing it in a hierarchical scheme, we insist on 
maintaining monophyletic groups and sister-group relationships. The discovery of 
monophyletic groups is the basic quest of phylogenetics. Going to all the trouble of finding 
the groups and then throwing them away does not make sense to us. 

Ever since the general theory of evolution gained acceptance, systematists have sought 
the one evolutionary history for organisms and have tried to fit that history into a hierarchical 
structure. We seek to reflect in our classifications the groups that we find in nature. Because 
phylogenetic reasoning delimits groups based on common ancestry, we can attempt to 
reconstruct evolutionary histories and from them develop a hierarchical ranking scheme. 
Phylogenetic groups are then a reflection of the order in nature. Therefore, our classifications 
can be used for the study of other characters and for further investigations in biogeography. 
coevolution, molecular evolution, rates of evolution, ecology, etc. If you wish to use 
classifications to study evolution, they must reflect the genealogy of the taxa in question. 
Groups that are potentially artificial cannot be used in such investigations. 

One of the greatest strengths of the phylogenetic system is that the method and results are 
transparent, meaning that decisions, whether right or wrong, are based on data that can be 
examined by any and all persons willing to understand the nature of the data. The 
phylogenetic system does not depend on some special and mysterious knowledge about 
organisms that only the "expert" can understand. A critic cannot claim that your idea of the 
phylogenetic history of a group is wrong just because he has smdied the groups longer than 
you have. Of course, there are valid disagreements, and there is room for change and 
improvement. But these disagreements are data based, not opinion based. Phylogenedcs, to 
put it crudely, is a put-up-or-shut-up scientific discipline. 

This workbook presents the basics of phylogenetic systematics as we use it today. We also 
cite references for those interested in following some of the debates currently underway 
among the proponents of phylogenetic systematics. We hope that this information will 
stimulate you and illustrate the importance of systematics as the basis of comparative 
biology. When you have finished this workbook, you should be able to reread this 
introduction and understand what we are trying to accomplish. As an acid test, go read 
Hennig (1966); it's the way we got started, and it remains the classic in the field. 

All new scientific ideas and analytical methods are accompanied by new sets of terms and 
concepts, which can be unsettling to the tyro and even more unsettling to the experienced 
systematist who is called upon to abandon the "traditional" meanings of terms and embrace 
new meanings. The basic rationale for adopting the definitions and concepts presented in 
this workbook is twofold. First, it is vitally important for systematics and taxonomy to be 
integrated into in the field of evolutionary theory. Willi Hennig's major motivation for 
reforming systematics and taxonomy was to bring them in line with the Darwinian 
Revolution, making the results obtained through phylogenetic systematics directly relevant 
to studies in other fields of evolutionary research. Second, it is vitally important that the 



INTRODUCTION. TERMS. AND CONCEPTS 3 

temis used in an empirical field be as unambiguous as possible so that hypotheses are as 
clear as possible. With these rationales in mind, we offer the following definitions for the 
basic terms in our field. They are largely taken from Hennig ( 1 966) or Wiley ( 1 980, 198 la). 
Other more specialized tenns will be introduced in other chapters. 

Terms for Groups of Organisms 

1 . Taxon. — A taxon is a group of organisms that is given a name. Tlie name is a proper 
name. The form of many of these proper names must follow the rules set forth in one of the 
codes that govern the use of names. The relative hierarchical position of a taxon in a 
classification can be indicated in many ways, hi the Linnaean system, relative rank is 
denoted by the use of categories. You should not confuse the rank of a taxon with its reality 
as a group. Aves is a taxon that includes exactly the same organisms whether it is ranked as 
a class, an order, or a family. 

2. Natural taxon. — A natural taxon is a group of organisms that exists in nature as a 
result of evolution. Although there are many possible groupings of organisms, only a few 
groupings comprise natural taxa. In the phylogenetic system, there are two basic kinds of 
natural taxa: species and monophyletic groups. A species is a lineage. It is a taxon that 
represents the largest unit of taxic evolution and is associated with an array of processes 
termed speciation. A monophyletic group is a group of species that includes an ancestral 
species and all of its descendants (Fig. 1.1a). Members of monophyletic groups share a set of 
common ancestry relationships not shared with any other species placed outside the group. 
In other terms, a monophyletic group is a unit of evolutionary history. Examples include 
Mammalia and Angiospermae. 

3. Clade. — A clade is a monophyletic group, i.e., a natural taxon. 

4. Ancestral taxon. — An ancestral taxon is a species that gave rise to at least one new 
daughter species during speciation, either through cladogenesis or reticulate speciation. By 
cladogenesis we mean speciation that results in two or more branches on the phylogenetic 
tree where there was only one branch before. By reticulate speciation we mean the 
establishment of a new species through a hybridization event involving two different 
species. A species that emerged from cladogenesis has one ancestral species but a species 
emerging from reticulate speciation has two ancestral species. In the phylogenetic system, 
only species can be ancestral taxa. Groups of species are specifically excluded from being 
ancestral to other groups of species or to single species. The biological rationale for this 
distinction is clear; there is an array of processes termed speciation that allow for one species 
to give rise to another (or two species to give rise to a species of hybrid origin), but there are 
no known processes that allow for a genus or a family to give rise to other taxa that contain 
two or more species ("genusation" and "familization" are biologically unknown). Thus, 
each monophyletic group begins as a single species. This species is the ancestor of all 
subsequent members of the monophyletic group. 

5. Artificial taxon. — An artificial taxon is one that does not correspond to a unit 
involved in the evolutionary process or to a unit of evolutionary history. You will encounter 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



a 







B C 




M 



N 



\ 


A 


3 


C 


\ 


\\ 


/ 


/ 



Fig. 1.1. — Examples of monophyletic (a), paraphyletic (b). and polyphyletic (c) groups. 



two kinds of artificial groups. Paraphyletic groups are artificial because one or more 
descendants of an ancestor are excluded from the group (Fig. 1.1b). Examples include 
Dicotyledonae, Vermes, and Reptilia. Polyphyletic groups are artificial because the 
common ancestor is placed in another taxon (Fig. 1.1c). An example would be the 
Homeothermia, a group composed of birds and mammals. Note that the term "ancestor" is 
used in its logical sense, i.e., the ancestor is unknown but its inclusion or exclusion can be 
deduced as a logical consequence of the grouping. The iinportant contrast is between 
monophyletic groups and nonmonophyletic groups. Paraphyletic groups are as artificial as 
polyphyletic groups. Further, it is not always possible to distinguish clearly the status of a 
group as either paraphyletic or polyphyletic. 

6. Grade. — A grade is an artificial taxon. Grade taxa are frequently paraphyletic and 
sometimes polyphyletic but are supposed to represent some level of evolutionary progress, 
level of organization, or level of adaptation (e.g., Reptilia or Vennes). 

7. Ingroup. — The ingroup is the group actually studied by the investigator (Fig. 1.2a). 
That is, it is the group of interest. 

8. Sister group. — A sister group is the taxon that is genealogically most closely related 
to the ingroup (Fig. 1.2a). The ancestor of the ingroup cannot be its sister because the 
ancestor is a member of the group. 



INTRODUCTION, TERMS, AND CONCEPTS 



2nd Outgroup 



Sister Group 



Ingroup 



a 




Branch 



Node 
Internode 



M 



B 



N 



Fig. 1.2. — A rooted (a) and unrooted (b) tree for the group ABC and two of its outgroups, N (the sister 
group) and M. 



9. Outgroup. — An outgroup is any group used in an analysis that is not included in the 
taxon under study. It is used for comparative purposes, usually in arguments conceming the 
relative polarity of a pair (or series) of homologous characters. The most important outgroup 
is the sister group, and considerable phylogenetic research may be needed to tind the sister 
group. Usually more than one outgroup is needed in an analysis. This will become apparent 
in Chapters. 



Quick Quiz — Groups 

Examine Fig. 1.1 and answer the following: 

1 . Why do we say that the group A+B+C and the group M+N are monophyletic? 

2. Which taxa would have to be either included or excluded to change the paraphyletic groups 
into monophyletic groups? 

3. Can polyphyletic groups ever contain monophyletic groups within them? 

4. Where are the ancestors in these diagrams? 



6 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 1 9 

Terms for the Relationships of Taxa 

1. Relationship. — In the phylogenetic system, the term relationship refers to the 
genealogic or "blood" relationship that exists between parent and child or between sister and 
brother. In other systems, relationship can also refer to similarity, with the evolutionary 
implication that taxa that are more similar to each other are more closely related. This 
meaning is specifically excluded fi'om the phylogenetic system. 

2. Genealogy and genealogic descent. — A genealogy is a graphic representation of the 
descent of offspring from parents. Genealogic descent on the taxon level (i.e., between 
groups recognized as taxa) is based on the proposition that species give rise to daughter 
species through an array of mechanisms termed speciation. 

3. Tree. — A tree is a branching structure and, in our sense, might contain reticulations as 
well as branches. A tree may be rooted (Fig. 1 .2a) or unrooted (Fig. 1 .2b) and is composed of 
several parts. A branch is a line connecting a branch point to a terminal taxon. A branch 
point, or node, represents a speciation event. This is true even if the taxa joined by the branch 
point are higher taxa such as families or phyla, because higher taxa originated as species. 
Branch points are sometimes represented by circles. An internode is a line connecting two 
speciation events and represents at least one ancestral species. (We say at least one because 
the statement is made relative to the species and groups we actually know about. It is always 
possible to find a new species or group of species that belongs to this part of the phytogeny. 
To make this addition, we would bisect the internode and create the possibility for an 
additional ancestral species.) The internode at the bottom of the tree is given the special term 
root. The term interval is a synonym of internode and is used in the Wagner algorithm (see 
Chapter 4). A neighborhood is an area of a tree relative to a particular taxon or taxa. In Fig. 
1.2b, taxon B is the nearest neighbor of taxa A and C. Note that A may or may not be the 
sister of a monophyletic group B+C. This relationship cannot be established until the root is 
specified. 

4. Phylogenetic tree. — A phylogenetic tree is a graphic representation of the genealogic 
relationships between taxa as these relationships are understood by a particular investigator 
In other words, a phylogenetic tree is a hypothesis of genealogic relationships on the taxon 
level. Although it is possible for an investigator to actually name ancestors and associate 
them with specific internodes, most phylogenetic trees are common ancestiy trees. Further, 
phylogenetic trees are hypotheses, not facts. Our ideas about the relationships among 
organisms change with increasing understanding. 

5. Cladogram. — Cladograms are phylogenetic trees. They have specific connotations of 
implied ancestry and a relative time axis. Thus, a cladogram is one kind of phylogenetic tree, 
a common ancestry tree. In some modifications of the phylogenetic system, specifically 
what some have termed Transformed Cladistics, the cladogram is the basic unit of analysis 
and is held to be fundamentally different from a phylogenetic tree. Specifically, it is purely a 
depiction of the derived characters shared by taxa with no necessary connotation of common 
ancestry or relative time axis. 



INTRODUCTION, TERMS, AND CONCEPTS 



6. Venn diagram. — A Venn diagram is a graphic representation of the relationships 
among taxa using intemested circles and ellipses. The ellipses take the place of intemode 
connections. A typical Venn diagram is contrasted with a phylogenetic tree in Fig. 1 .3. 



Aves 



Crocodylia Lepidosauria 



a 




f Aves J f Crocodylia j f Lepidosauria j 



Fig. 1 .3. — A phylogenetic tree (a) and a Venn diagram (b) of three groups of tetrapod vertebrates. 



Quick Quiz — Relationships 

Examine Fig. 1.2a and answer the following: 

1 . What is the sister group of the clade N? 

2. What is the sister group of the clade M? 

3. What is the sister group of a group composed of M+N? 

4. Where is the hypothetical ancestor of the ingroup on the tree? 

5. How many ancestors can a group have? 

6. Draw a Venn diagram of Fig. 1 .2a. 



Terms for Classifications 

1 . Natural classification. — A classification containing only monophyletic groups and/or 
species is a natural classification. A natural classification is logically consistent with the 
phylogenetic relationships of the organisms classified as they are understood by the 
investigator constructing the classification. That is, the knowledge claims inherent in a 
natural classification do not conflict with any of the knowledge claims inherent in the 
phylogenetic tree. The protocols for determining if a classification is logically consistent 
with a phylogenetic tree are given in Chapter 6. 



8 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

2. Artificial classification. — An artificial classification is a classification containing one 
or more artificial groups (i.e., one or more paraphyletic or polyphyletic groups). An artificial 
classification is logically inconsistent with the phylogenetic relationships of the organisms 
as they are understood by the investigator making the classification. That is, some of the 
knowledge claims inherent in the classification conflict with knowledge claims in the 
phylogenetic tree. 

3. Arrangement. — An arrangement is a classification of a group whose phylogenetic 
relationships are not known because no investigator has ever attempted to reconstruct the 
evolutionary history of the group. The vast majority of current classifications are 
arrangements. A particular arrangement may turn out to be either a natural or an artificial 
classification. Arrangements serve as interim and completely necessary vehicles for 
classifying organisms until the phylogenetic relationships of these organisms can be worked 
out. 

4. Category. — The category of a taxon indicates its relative place in the hierarchy of the 
classification. The Linnaean hierarchy is the most common taxonomic hierarchy and its 
categories include class, order, family, genus, and species. The formation of the names of 
taxa that occupy certain places in the hierarchy are governed by rules contained in various 
codes of nomenclature. For example, animal taxa ranked at the level of the categoiy family 
have names that end in -idae, whereas plant taxa ranked at this level have names that end in 
-aceae. It is important to remember that the rank of a taxon does not affect its status in the 
phylogenetic system. To the phylogeneticist, all monophyletic taxa are equally important 
and all paraphyletic and polyphyletic taxa are equally misleading. 

Classifications and arrangements are usually presented as hierarchies of names, with 
relative position in the hierarchy (rank) noted by categories. However, these classifications 
can be portrayed as tree diagrams and as Venn diagrams. The use of these methods of 
presenting classifications is discussed in Chapter 6. 



Quick Quiz — Classification 

1. In the phylogenetic system, must the taxa be clades? 

2. In the phylogenetic system, must categories be clades? 

3. Which is more important, a phylum or a genus? 



Process Terms 

Three process terms are of particular importance in the phylogenetic system. Speciation 
results in an increase in the number of species in a group. Speciation is not a single process 
but an array of processes. Cladogenesis is branching or divergent evolution and is caused by 
speciafion. Anagenesis is change within a species that does not involve branching. The 
extent to which anagenesis and cladogenesis are coupled is an interesting evolutionary 
question but not a question that must be settled to understand the phylogenetic system. 



INTRODUCTION. TERMS, AND CONCEPTS 9 

Terms for the Attributes of Specimens 

1 . Character. — A character is a feature, that is, an observable part of, or attribute of, an 
organism. 

2. Evolutionary novelty. — An evolutionary novelty is an inherited change from a 
previously existing character. The novelty is the homologue of the previously existing 
character in an ancestor/descendant relationship. As we shall see below, novelties are 
apomorphies at the time they originate. 

3. Homologue. — Two characters in two taxa are homologues if one of the following two 
conditions are met: 1) they are the same as the character that is found in the ancestor of the 
two taxa or 2) they are different characters that have an ancestor/descendant relationship 
described as preexisting/novel. The ancestral character is temied the plesiomorphic 
character, and the descendant character is termed the apomorphic character. The process 
of detennining which of two homologues is plesiomorphic or apomorphic lies at the heart of 
the phylogenetic method and is termed character polarization or character 
argumentation. Three (or more) characters are homologues if they meet condition 2. 

4. Homoplasy. — A homoplasy is a similar character that is shared by two taxa but does 
not meet the criteria of homology. Every statement of homology is a hypothesis subject to 
testing. What you thought were homologues at the beginning of an analysis may end up to be 
homoplasies. 

5. Transfonnation series. — A transformation series (abbreviated TS in some tables and 
exercises) is a group of homologous characters. If the transfonnation series is ordered, a 
particular path of possible evolution is specified but not necessarily the direction that path 
might take. All transformation series containing only two homologous characters (the binary 
condition) are automatically ordered but not necessarily polarized (contrast Fig. 1.4a and 
Fig. 1 .4b). Transfonnation series having more than two characters are termed multicharacter 
(or multistate) transformation series. If a multistate transformation series is unordered (Fig. 
1 .4c), several paths might be possible. Ordered transformation series are not the same as 
polarized transfonnation series (compare Figs. 1.4d and 1.4e). An unpolarized 
transformation series is one in which the direction of character evolution has not been 
specified (Figs. 1 .4a, c, d). A polarized transformation series is one in which the relative 
apomorphy and plesiomorphy of characters has been determined by an appropriate criterion 
(Figs. 1 .4b, e). It is possible for a transformation series to be both unordered and polarized. 
For example, we might know from outgroup comparison that is the plesiomorphic state, 
but we might not know whether 1 gave rise to 2, or vice versa, or whether 1 and 2 arose 
independently from 0. Ordering and polarization of multicharacter transformation series can 
become very complicated, as we shall see in Chapter 3. Our use of the convention 
"transformation series/character" differs from that of many authors who use "character" as a 
synonym for "transfonnation series" and "character state" as a synonym for "character." We 
use "transfonnation series/character" instead of "character/character state" in our research 
and in this workbook for philosophical reasons. The "character/character state" convention 
reduces "character" to a term that does not refer to the attributes of organisms but instead to 
a class construct that contains the attributes of organisms, homologues or not. For example. 



10 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 1 9 



a -< >- 1 

b >- 1 C 1 -< >■ 2 



/\ 



-< >► 1 -^r- 



-*► 1 > 2 



Fig. 1.4. — Characters, a. Unpolarized binary characters, b. Polarized binary characters, c. An unordered 
transformation series of three characters, d. The same transformation series ordered but not polarized, e. The 
same transformation series ordered and polarized. 



dandelions do not have "color of flower" as an attribute; they have "yellow flowers." We 
adopt "transformation series/character" because it explicitly avoids the construction of 
character classes and implicitly encourages the investigator to use characters hypothesized 
to be homologues of each other. 

6. Character argumentation. — Character argumentation is the logical process of 
determining which characters in a transformation series are plesiomorphic and which are 
apomorphic. Character argumentation is based on a priori arguments of an "if, then" 
deductive nature and is based on outgroup comparison. This process is frequently termed 
"polarizing the characters." Polarity refers to which of the characters is plesiomorphic or 
apomorphic. Character argumentation will be covered in detail in Chapter 3. 

7. Character optimization. — Character optimization consists oi a posteriori arguments 
as to how particular characters should be polarized given a particular tree topology. 
Character optimization might seem a priori when used in a computer program, but it is not. 

8. Character code and data matrix. — Phylogenetic systematists are quickly converting to 
computer- assisted analysis of their data. When using a computer, the investigator produces a 
data matrix. Usually, the columns of the matrix are transformation series and the rows are 
taxa. A code is the numerical name of a particular character. By convention, the code " 1 " is 
usually assigned to the apomorphic character and "0" to the plesiomorphic character of a 
transformation series if the polarity of that series is determined (=hypothesized) by an 
appropriate method of polarization. If the transformation series consists of more than two 
characters, additional numerical codes are assigned. Alternatively, the matrix might be 
coded using binary coding as discussed in Chapter 3. There are many ways of reflecting the 
code of a character when that character is placed on a tree. We will use the following 
convention: characters are denoted by their transformation series and their code. The 
designation 1-1 means "transfomiation series 1, character coded 1." Some basic ways of 
coding characters are discussed in Chapter 3. 



INTRODUCTION, TERMS, AND CONCEPTS 1 1 

9. Tree length. — The length of a tree is usually considered the number of evolutionary 
transformations needed to explain the data given a particular tree topology. 

You will probably need some time to assimilate all of the definitions presented. A good 
strategy is to review this chapter and then go to Chapter 2. working your way through the 
examples. We have found that deeper understanding comes from actual work. Although we 
cannot pick a real group for you to work on, we have attempted the next best thing, a series 
of exercises designed to teach basic phylogenetic techniques that we hope will mimic real 
situations as closely as possible. 



Quick Quiz — Cliaracters 

1 , How would the transformation series in Fig. 1 .4c look if it were polarized and unordered? 

2. Is character "l" in Fig. 1 .4e apomorphic or plesiomorphic? 



Chapter Notes and References 

1. There is no substitute for reading Hennig (1966). We suggest, however, that you 
become familiar with most of the basics before attempting to read the 1966 text. Hennig 
(1965) is the most accessible original Hennig. Other classics include Brundin (1966) and 
Crowson ( 1970). An interesting analysis of Hennig's impact on systematics can be found in 
Dupuis (1984). A considerable portion of the history of phylogenetic thought (and indeed 
post- 1950 systematics) can be followed in a single journal, Systematic Zoology. We highly 
recommend that students examine this journal. 

2. Post-Hennig texts that are suitable for beginners are Eldredge and Cracraft (1980), 
Wiley (1981a). Ridley (1985), Schoch (1986), Ax (1987), and Sober (1988a). A more 
difficult text written from the point of view of the transformed cladists is Nelson and 
Platnick(1981). 

3. A very readable review of the entire field of systematics is Ridley (1985), whose 
defense of phylogenetics and criticisms of traditional (evolutionary) taxonomy, phenetics, 
and transformed cladistics are generally on the mark. 

Quick Quiz Answers 
Groups 

1 . They are monophyletic because no descendant of their respective common ancestor is left out of the 
group. 

2. To make the group 0+A+B monophyletic, you would have to include C. To make the group 
N+O+A+B+C monophyletic. you could either include M or exclude N. 

3. Yes; e.g., N+A+B+C contains the monophyletic group ABC. 

4. You were pretty clever if you answered this one because we haven't covered it yet. The ancestors are 
represented by intemodes between branches. Obviously they are hypothetical because none of them are 
named. 



12 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



Relationships 

1 . The ingroup (A+B+C) is the sister group of N. 

2. N plus the ingroup is the sister group of M. 

3. A group composed of M and N is paraphyletic. Paraphyletic groups are artificial and thus cannot have 
sister groups. 

4. The intemode labeled "Intemode." 

5. A bunch, stretching back to the origin of life. But we usually refer only to the immediate common 
ancestor. 

6. 



M 




Classification 

1. Only clades (monophyletic groups and species) are permitted in the phylogenetic system. Grades are 
specifically rejected. As you will see in Chapter 6, this is because classifications that contain even a single 
grade are logically inconsistent with the phylogeny of the group containing the grade. 

2. Categories are not taxa. They are designations of relative rank in a classification. As such, categories are 
neither clades nor grades. 

3. All monophyletic taxa are equally important and interesting to the phylogeneticist. 

Characters 




1 -^ 



■^ 2 



2. Character 1 is both apomorphic and plesiomorphic. It is apomorphic relative to and plesiomorphic 
relative to 2. 



Chapter 2 



BASIC PHYLOGENETIC TECHNIQUES 



Phylogenetic systematists work under the principle that there is a single and historically 
unique genealogic history relating all organisms. Further, because characters are features of 
organisms, they should have a place on the tree representing this history. The proper place 
for a character on the tree is where it arose during evolutionary history. A "proper" tree 
should be one on which the taxa are placed in correct genealogic order and the characters are 
placed where they arose. For example, in Fig. 2.1 we show a tree of some major land plant 
groups with some of their associated characters. This tree can be used to explain the 
association of characters and taxa. Tlie characters xylem and phloem are placed where they 
are because the investigator has hypothesized that both arose in the common ancestor of 
mosses and tracheophytes. In other words, they arose between the time of origin of the 
homworts and its sister group. Xylem and phloem are thought to be homologous m all plants 



Liverworts Homworts 



Mosses Tracheophytes 



Oil bodies 
Elaters in 
sporangium 
Lunulanc acid 



Pseudo-elaters 
in sporangium 

- - Spore production 
nonsynchronous 

- - Intercalary meristem 
in sporophyte 



Leaves on 

gametophyte 
Multicellular 

rhyzoids 



-- True lignin 

-- Ornamented tracheid 
walls 

Independent sporophyte 
Branched sporophyte 



Xylem 

Phloem 

Polyphenolics in xylem wall 

Perine layer on spores 

Aerial sporophyte axis 




- - Ability to distinguish o-methionine 
-- Stomates 




Fig. 2.1. — The phylogenetic relationships among several groups of plants (after Bremer, 1985) 
Synapomorphies and autapomorphies for each group are listed. Some characters are not shown. 



14 KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 

that have these tissues. Thus, each appears only once, at the level of the tree where each is 
thought to have arisen as an evolutionary novelty. Now, if we did not have this tree but only 
the characters, we might suspect that all of the taxa that have xylem and phloem shared a 
common ancestor that was not shared by taxa that lack xylem and phloem. Our suspicion is 
based on a complex of hypotheses, including the fact that xylem and phloem from different 
plants are very similar, being composed of a few basic cell types or obvious derivatives of 
these cell types, hi the phylogenetic system, such detailed similarity is always considered 
evidence for homology, which necessarily implies common origin. This concept is so 
important that it has been given the name Hennig's Auxiliary Principle (Hennig, 1953, 1966). 

Hennig's Auxiliary Principle. — Never assume convergence or parallel evolution, always 
assume homology in the absence of contrary evidence. 

This principle is a powerful one. Without it, we must pack up and go home because we 
could exclude any similarity that we don't wish to deal with by asserting that "it probably 
arose by convergent evolution." Of course, just because we use Hennig's Auxiliary Principle 
doesn't mean that we believe that convergences are rare or nonexistent. Convergences are 
facts of nature and are rather common in some groups. But to pinpoint convergence, you first 
have to have a tree, and without Hennig's Auxiliary Principle you will never get to a tree 
because you will be too worried that the characters you study are convergences. Hennig is 
simply suggesting that you sit back and let his method worry about convergences rather than 
doing something rash and ad hoc. 

Back to the xylem and phloem. With Hennig's Auxiliary Principle, you can deduce that 
plants that have xylem and phloem shared a common ancestor not shared with other plants. 
Of course, you don't make such a deduction in a vacuum. You "know" that more primitive 
plants lack xylem and phloem, and thus it's a good guess that having xylem and phloem is 
more derived than lacking xylem and phloem. This deduction is a primitive sort of outgroup 
comparison, which we will discuss in some detail in Chapter 3. For now, we want you to 
consider another principle. 

Grouping Rule. — Synapomorphies are evidence for common ancestry relationships, 
whereas symplesiomorphies, convergences, and parallelisms are useless in providing 
evidence of common ancestry (Hennig, 1966). 

Intuitively, you know convergences and parallelisms (both termed homoplasies) are 
useless for showing common ancestry relationships because they evolved independently. 
However, plesiomorphies are also homologies. So, why can't all homologies show common 
ancestry relationships? The answer is that they do. It's just that symplesiomorphies can't 
show common ancestry relationships at the level in the hierarchy you are working at because 
they evolved earlier than any of the taxa you are trying to sort out. In addition, they have 
already been used at the level where they first appeared. If they hadn't been used, you would 
not be where you are. For example, you would never hypothesize that pineapples are more 
closely related to mosses than to some species of mistletoe based on the plesiomorphy 
"presence of chlorophyll." If you accepted that as valid evidence, then you would have to 



BASIC PHYLOGENETIC TECHNIQUES 15 

conclude that pineapples are also more closely related to green algae than to mistletoes. A 
common complaint by traditional taxonomists is that "cladists" use only part of the data 
available to them (Cronquist, 1987). This is not true, as the above example demonstrates. 
What we do is to attempt to find the correlation between the relative age and origin of 
characters. 

Finally, we have to consider how to combine the information from different 
transformation series into hypotheses of genealogic relationships. Tliere are several ways of 
accomplishing this, depending on the algorithm you use. We will find out more about this as 
we proceed through the workbook. For now, we will use an old-fashioned (and perfectly 
valid) grouping rule that goes back to the roots of the phylogenetic method, the inclusion/ 
exclusion rule. This rule is implicit in the early work of Hennig ( 1 966), as well as being used 
as an explicit rule in the much later group compatibility algorithm developed by M. Zandee 
(Zandee and Geesink, 1987). 

Inclusion/Exclusion Rule. — ^The information from two transformation series can be com- 
bined into a single hypothesis of relationship if that infomiation allows for the complete 
inclusion or the complete exclusion of groups that were formed by the separate transfor- 
mation series. Overlap of groupings leads to the generation of two or more hypotheses of 
relationship because the infomiation cannot be directly combined into a single hypothesis. 

The inclusion/exclusion rule is directly related to the concept of logical consistency. Trees 
that conform to the rule are logically consistent with each other. Those trees that show 
overlap are logically inconsistent with each other. This can be shown graphically using Venn 
diagrams. 

You can get an idea of how this rule works by studying the examples in Fig. 2.2. In Fig. 
2.2a, we have four characters and four trees. The first tree contains no character information. 
It is logically consistent with any tree that has character information. The second tree states 
that N, O, and P form a monophyletic group based on characters from two transformation 
series (1-1 and 2-1 ). Tlie third tree states that O and Pfomi a monophyletic group based on 
two additional characters (3-1 and 4-1 ). Note that 0-i-P is one of the possible groupings that 
could be found in the group N-i-O-i-P, and N-i-O+P completely includes 0+P. Tlie fourth tree 
combines these logically consistent hypotheses of relationship. Thus, these data lead to two 
groupings that are logically consistent with each other. The second example. Fig. 2.2b, 
shows the result of the inclusion of two smaller monophyletic gioups (S-i-T) and (U-i-V) 
within a larger group (S-V). In Fig. 2.2c, we have an example of the violation of the 
inclusion/exclusion rule. All six transfonnation series imply groupings that can be included 
within the larger group A-D. Both C-i-B and C-t-D can be included within the group B-i-C-i-D. 
However, their knowledge claims conflict, and the groups overlap (Fig. 2. 2d). 
Transformation series 1 - 1 and 2- 1 imply a group C-hB while excluding D, and transfomiation 
series 3-1 and 4-1 imply a group C-i-D while excluding B. C is included in two different 
groups, as shown by the Venn diagram in Fig. 2. 2d. As a result, there are two equally 
parsimonious trees that are logically inconsistent with each other. To resolve which of these 
trees (or another tree) is preferable, we would have to analyze more data. 



16 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



MNOP M NOP MNOP 



NOP 






STUV STUV UVST 







ABC D 



•4-1 

ADCB ABCD ABCD \ /g.^{^ 

■5-1 








r 




A 




B (c 


°) 








> 




V 





Fig. 2.2. — ^Three examples (a-c) of the use of the inclusion/exclusion rule for combining the information 
of different transformation series into trees, d. A Venn diagram showing the logical inconsistency in c. 



BASIC PHYLOGENETIC TECHNIQUES 17 



Quick Quiz — Basic Rules of Analysis 

Does it follow from Hennig's Auxiliary Principle that birds and insects share a common 

ancestor not shared with, say, crocodiles because both have wings and are capable of flight? 

Lizards and crocodiles have amniotic eggs. Does it follow from the Grouping Rule that 

lizards and crocodiles share a common ancestor? 

How can you tell that "presence of chlorophyll a" is a plesiomorphy rather than a 

synapomorphy? 



Sample Analyses 

We will cover the complexities of character argumentation in the next few chapters. The 
exercises below are based on the proposition that the outgroup has plesiomorphic characters. 
You can determine which character of the transformation series is plesiomorphic by simple 
inspection of the outgroup. (By the time you get through with Chapter 3, you will see that 
such a simple rule doesn't always hold, but it's good enough to get through these exercises.) 
Grouping is accomplished by application of the Grouping Rule. We will first take you 
through two exercises. Then we present a series of data matrices for you to work with. 
(Solutions to all exercises are in the back.) 

Example 2.1. — The relationships of ABCidae. 

1 . Examine transformation series (TS) 1 in Table 2. 1 . It is composed of characters in the 
first column of the data matrix. We can draw a tree with the groupings implied by the 
synapomorphy found in the transformation series. We can do the same for TS 2. Our results 
look like the two trees to the left in Fig. 2.3. Because both imply the same groupings, we can 
say that they are topologically identical. That is, they are isomorphic. The combination of 
the two trees, by applying the Grouping Rule, is the tree on the right. We can calculate a tree 
length for this tree by simply adding the number of synapomorphies that occur on it. In this 
case, the tree length is two steps. 



Table 2. 1 . — Data matrix for ABCidae (Example 2. 1 ). 











Transformation 


series 








Taxon 


1 


2 


3 


4 




5 


6 


7 


X (outgroup) 

























A 


1 


1 



















B 


1 


1 


1 


1 













C 


1 


1 


1 


1 




1 


1 


1 



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KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



B C 



X 



B 



X 



B C 






Fig. 2.3. — Trees for transformation series 1 and 2 (Example 2. 1 ). 

2. If we inspect TS 3 and TS 4, we see that the synapomorphies have identical distribu- 
tions, both implying that B and C form a monophyletic group (Fig. 2.4). If we put both of 
these synapomorphies on a tree, the results should look like the tree on the right in Fig. 2.4. 
What is the length of this tree? 




X 



B 



X 



B 





Fig. 2.4. — Trees for transformation series 3 and 4 (Example 2.1). 

3. Note that only C has the apomorphies in TS 5, 6, and 7. Unique (single occurrence) 
apomorphies are termed autapomorphies. They are not useful for grouping, as we can see 
in Fig. 2.5, but they are useful for diagnosing C. Autapomorphies also count when figuring 
tree length but not when comparing trees. The length of this tree is three steps. 



X 



B 



C 




Fig. 2.5. — ^Tree for transformation series 5, 6, and 7 (Example 2. 1 ). 



4. We now have three different tree topologies. If we look at them closely, we can see that 
although the three trees are topologically different, they do not contain any conflicting 
information. For example, the tree implied by TS 5-7 does not conflict with the trees implied 
by the other transformation series because all that TS 5-7 imply is that C is different from the 



BASIC PHYLOGENETIC TECHNIQUES 



19 



Other four taxa. Further, TS 1 and 2 do not conflict with TS 3 and 4 because TS 1 and 2 imply 
that A, B, and C form a monophyletic group, whereas TS 3 and 4 imply that B and C form a 
monophyletic group but say nothing about the relationships of A or the outgroup, X. Trees 
that contain different but mutually agreeable groupings are logically compatible or fully 
congruent. They can be combined without changing any hypothesis of homology, and the 
length of the resulting tree is the sum of the lengths of each subtree. For example, we have 
combined all of the information in the data matrix to produce the tree in Fig. 2.6. Its length is 
seven steps, the total of the number of steps of the subtrees. 



X 



B 




Fig. 2.6. — The best estimate of the common ancestry relationships of A, B, and C, given the data in 
Example 2.1. 



Example 2.2. — Analysis of MNOidae. 

The first thing you should notice about this matrix (Table 2.2) is that it has more characters 
scored as "1 ." Let's work through it. 



Table 2.2. — Data matrix for MNOidae (Example 2.2). 











Transformation 


series 








Taxon 


1 


2 


3 


4 




5 


6 


7 


X (outgroup) 

























M 


1 


I 










1 


1 


1 


N 


1 


1 


1 


1 




1 


1 


1 





1 


1 


1 


1 














1 . TS 1 and TS 2 imply that M, N, and O form a monophyletic group as shown in Fig. 2.7. 

X M N O 




Fig. 2.7. — Tree for transformation series 1 and 2 (Example 2.2). 



20 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



2. TS 3 and TS 4 imply that N and O form a monophyletic group (Fig. 2.8). 



X M N O 




Fig. 2.8. — Tree for transformation series 3 and 4 (Example 2.2). 



3. Finally, TS 5, 6, and 7 imply that M and N form a monophyletic group (Fig. 2.9). 



X 



o 



N 



M 




Fig. 2.9. — Tree for transformation series 5, 6, and 7 (Example 2.2). 

4. At this point you should suspect that something has gone wrong. TS 3 and 4 imply a 
monophyletic group that includes N and O but excludes M, whereas TS 5-7 imply a 
monophyletic group that includes M and N but excludes O. There must be a mistake, 
because we have violated the inclusion/exclusion rule. In such a situation, we invoke 
another important principle of phylogenetic analysis: there is only one true phylogeny. 
Thus, one of our groupings must be wrong, (hi fact, they might both be wrong, but the 
Auxiliary Principle keeps us going until such time that we demonstrate that both are wrong.) 
In this situation, we are faced with two logically incompatible trees (Fig. 2.10). Note that 
there is some congruence because both trees have the apomorphies of the first two 
transformation series. 





Fig. 2. 10. — ^Trees for the two different sets of consistent transformation series (Example 2.2). 



BASIC PHYLOGENETIC TECHNIQUES 



21 



5. You should have guessed by now that neither of the trees shown above is really a 
complete tree. The tree on the right lacks TS 5, 6, and 7, and the one on the left lacks TS 3 and 
4. Leaving out characters is not acceptable. About the only way that you can get into more 
trouble in phylogenetic analysis is to group by symplesiomorphies. Before we start, consider 
how characters might be homoplasious. A character might be a convergence/parallelism, or 
it might be a reversal to the "plesiomorphic" character We must consider both kinds of 
homoplasies. hi Fig. 2. 1 1 a, TS 3 and 4 are put on the tree under the assumption that 3- 1 and 
4-1 arose independently (i.e., via convergence/parallelism). InFig. 2.11b, we have placed 5-1, 
6-1, and 7-1 on the alternate tree as convergences. In Fig. 2.1 Ic, we assume that 3-1 and 4-1 
arose in the common ancestor of the group and that M has reverted to the plesiomorphic 
character. Thus, 3-0 and 4-0 appear on the tree as autapomoiphies of M. We have done the 
same thing for O in Fig. 2. 1 Id for TS 5-7, given the alternative hypothesis. 







Fig. 2.11 . — Alternative hypotheses of the relationships of M, N. and O based on characters of Example 
2.2. ® = character showing convergence/parallelism or reversal (homoplasies). 



6. The question is — which of these trees should we accept? That turns out to be a rather 
complicated question. If we adhere to the Auxiliary Principle, we should strive for two 
qualities, the greatest number of homologies and the least number of homoplasies. These 
qualities are usually consistent with each other; that is, the tree with the greatest number of 
synapomorphies is also the tree with the least number of homoplasies. But you can find 



22 KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 

exceptions to this. Fortunately, both of these qualities are related to tree length. When you 
count the number of steps in the four trees in Fig. 2. 1 1 . you will find that trees a and c have 
nine steps, and trees b and d have 10 steps. We accept trees a and c as the best estimates of the 
phylogeny because they have a shorter length and thus the greatest number of statements of 
homology and the fewest number of statements of homoplasy for the data at hand. Note that 
such statements are relative only to trees derived from the same data set. The Auxiliary 
Principle coupled with the principle that there is only one phylogeny of life carried us to this 
point. Methodologically, we have employed the principle of parsimony. In the 
phylogenetic system, the principle of pai'simony is nearly synonymous with the Auxiliary 
Principle. We can see three additional characteristics. First, trees a and c are topologically 
identical. Therefore, the common ancestry relationships hypothesized are identical. Second, 
these two trees make different claims concerning character evolution. Third, they are 
equally parsimonious, therefore we cannot make a choice about character evolution unless 
we employ some parsimony criterion other than tree length. 

7. Finally, we can evaluate the performance of each character originally coded as a 
synapomorphy by calculating a consistency index for it. The consistency index (CI) of a 
character is simply the reciprocal of the number of times that a character appears on the tree. 
The CI is a favorite summary "statistic" in computer programs such as PAUP (Swofford, 
1990) and MacClade (Maddison and Maddison, in press); therefore, it is good to practice 
some hand calculations so that you will know how the CI works. We will discuss this index 
and other measures of tree comparisons in Chapter 5. For example, in one most parsimoni- 
ous tree (Fig. 2. 1 1 a), the apomorphy coded 1 in TS 3 appears twice, so its CI is 

CI = ^ = 0.5. 

2 

For a given tree, we can see that a character is not really a synapomorphy by simple 
inspection of its CI. True homologues (real synapomorphies) have CIs of 1.0. Of course, our 
best estimates of true homologues come a posteriori , that is, in reference to the best estimate 
of common ancestry relationships. We do not know in advance that a particular derived 
similarity will turn up with a CI less than 1 .0. 



Exercises 

For each of the exercises below do the following: 

1 . Derive trees for each transformation series or each set of transformation series with the 
same distribution of synapomorphies (like TS 1 and TS 2 in Example 2. 1 ). 

2. Combine the logically consistent subtrees into the shortest tree or trees accounting for 
aU of the transformation series. Don't forget to account for the homoplasies as well as the 
synapomorphies and autapomoiphies. In some data matrices, there will only be one such 
tree, in others there will be two. Tip: Search the trees obtained above for groups that reoccur. 
Use these first (for example, sus+tus and vus+uus in Exercise 2.1). 



BASIC PHYLOGENETIC TECHNIQUES 



23 



3. Calculate the length of each tree and the CI for each character originally coded as a 
synapomorphy. 

EXERCISE 2.1.— Analysis of Sus (Table 2.3). 

Table 2.3. — Data matrix for analysis of Sus (Exercise 2.1). 













Transformation 


series 










Taxon 


1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


11 


12 


Outgroup 






































S. sus 







1 


1 


1 


1 


1 





1 











S. tus 




1 


1 


1 


1 


1 











1 








S. uus 







1 


1 


























S. riis 




1 














I 


1 








1 





S. wus 




1 














1 


1 











1 



EXERCISE 2.2.— Analysis of Midae (Table 2.4). 

Table 2.4. — Data matrix for analysis of Midae (Exercise 2.2). 













Transformation 


series 










Taxon 


1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


11 


Outgroup 



































Mus 


1 







1 


1 


1 








1 


1 


1 


Nus 


1 










1 


1 


1 








1 


1 


Ous 


1 










1 


1 


1 








1 


1 


Pus 







1 


1 











1 











Qus 


1 




1 


1 





1 


1 














Rus 


1 




1 


1 








1 















EXERCISE 2.3.— Analysis of Aus (Table 2.5). 



Table 2.5. — Data matrix for analysis of Aus (Exercise 2.3). 













Transformation series 










Taxon 


1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


Outgroup 
A. aus 











1 






o" 

1 



1 


















A. bus 







1 





1 





1 











A. cus 




1 


1 




















1 


A. dus 




1 





1 











1 








A. eus 




1 





1 








1 





1 






24 KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 

Chapter Notes and References 

1 . All of the texts cited in Chapter 1 cover the fundamentals of reconstructing phyloge- 
netic relationships, but they each do so from a slightly to very different point of view. The 
inclusion/exclusion criterion is not usually seen in the form we present it. Indeed, it is not the 
way one usually goes about doing phylogenetic reconstructions. We adopted our approach 
because it seemed to be the simplest one to use to teach basic principles. The inclusion/ 
exclusion approach is explicit in the "group compatibility" approach of Zandee and Geesink 
(1987). 

2. Considerable controversy surrounds the philosophical nature of phylogenetic 
hypothesis testing and its relationship to parsimony. Farris (1979, 1983), Sober (1983), 
Kluge (1984, 1989), and Brooks and Wiley (1988) present parsimony as the relevant 
criterion for judging competing hypotheses. This is in direct contrast to Wiley (1975, 
1981a), who attempted to reconcile parsimony and the hypothetico-deductive approach of 
Popper (1965), or to Felsenstein (1978, 1983), who argued that parsimony might not be the 
preferred criterion. Most of this controversy has been summarized by Sober (1988a). 

Quick Quiz Answers 

1. Not when you examine the wings in detail. Insect wings have a completely different structure when 
compared with bird wings. They are so different that the best hypothesis is that the wings are not homologous. 
This leads to the hypothesis that flight has evolved independently in each group. Further, there is much 
evidence in the fomi of other characters that leads to the hypothesis that flying insects are more closely related 
to other insects that do not have wings and that birds are more closely related to other vertebrates. 

2. Yes, it does follow. However, many other vertebrate groups, such as birds and mammals, also have an 
amniotic egg. Thus, while lizards and crocodiles certainly share a common ancestor, we cannot hypothesize 
that they share a common ancestor not shared also with birds and mammals. Rather, the amniotic egg is a 
character that provides evidence for a common ancestry relationship among all amniotes (as a 
synapomorphy), and its presence in lizards and crocodiles is a plesiomorphic homologous similarity. 

3. You can deduce that the presence of chlorophyll a is a plesiomorphy in the same manner that you can 
deduce whether a character is an apomorphy, by outgroup comparison. This is covered in the next chapter. 



Chapters 

CHARACTER ARGUMENTATION AND CODING 



This chapter is designed to teach the following skills: 1 ) interpretation of a phylogenetic 
tree in terms of nodes and intemodes, 2) polarization of characters at nodes and intemodes 
on the phylogenetic tree according to the criterion of phylogenetic parsimony as evidenced 
by outgroup comparison, and 3) character coding. 

OuTGROUP Comparison 

Hennig ( 1966) and Brundin (1966) characterized the essence of phylogenetic analysis as 
the "search for the sister group." They recognized that if you can find the closest relative or 
relatives of the group you are working on, then you have the basic tools for deciding which 
characters are apomorphic and which are plesiomorphic in a transformation series. The 
argument goes something like this. As an investigator, you see that members of your group 
have two different but homologous characters, "round pupils" and "square pupils." As a 
phylogeneticist, you know that one of these characters, the apomorphic one, might diagnose 
a monophyletic group, but both cannot (the Grouping Rule). If you think about it, Hennig 's 
reasoning becomes clear. If you find square pupils in the sister group of the taxon you are 
studying, then it is fairly clear that "square pupils" is older than "round pupils," and if this is 
true, then "square pupils" must be the plesiomorphic character in the transformation series. 
Therefore, reasoned Hennig, the characteristics of the sister group are vital in making an 
intelligent decision regarding polarity within the taxon studied. The simplest rule for 
determining polarity can be stated in the following way. 

Rule for Determining Relative Apomorphy. — Of two or more homologous characters 
found within a monophyletic group, that character also found in the sister group is the 
plesiomorphic character, and the one(s) found only in the ingroup is (are) the apomorphic 
one(s). 

As it tums out, actual polarity decisions can be a little more complicated than our simple 
example. What if, for example, we don't know the exact sister group but only an array of 
possible sister groups? What if the sister group is a monophyletic group, and it also has both 
characters? What if our group is not monophyletic? What if "square pupils" evolved in the 
sister group independently? 

Polarity Decisions 

The answers to these questions depend on our ability to argue character polarities using 
some formal rules. The most satisfactory discussion of these rules was published by 
Maddison et al. (1984). We will present the case developed by them for situations where the 



26 



KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 



phylogenetic relationships among the outgroups and the relationships of these taxa to the 
ingroup are known. Some handy shortcuts, such as the "doublet rule." will also be covered. 
However, before we can examine these rules, we need to learn some terms. 

The ingroup node (IG node) represents the ancestor of the taxon we will eventually be 
analyzing, once we determine the polarities of our transformation series. The same 
characters are assigned to the IG node as would be assigned to the ancestral intemode, and 
the terms are interchangeable. If we determine that "square pupils" is primitive, then "square 
pupils" is assigned to the IG node. To know what is plesiomorphic within the group, we only 
need to know what character was found at the IG node. In this case, because "square pupils" 
is assigned to the IG node, the character "round pupils" can be used to diagnose a possible 
monophyletic group within the IG. Unfortunately, the ancestor isn't around, and thus we 
must infer what character it had. In Fig. 3.1, the character "square pupils" is coded "a," and 
"round pupils" is coded "b." The outgroup node (OG node) is the node immediately below 
the IG node. Don't be confused (like some of our students) and think that the OG node refers 
to characters that are associated with the IG intemode; instead, the OG node is associated 
with characters of the intemode immediately below it (in a manner similar to the IG node). 
So. the general rule is the internode is associated with the node directly above it. This is 
important because characters are usually put on intemodes rather than nodes; therefore, it is 
important to remember that these characters belong to the node above them rather than the 
node below them. 

Characters are designated by small letters and are placed where taxa are usually labeled. 
Letters are used purely as a heuristic device and to avoid connotations of primitive and 



OG 



a 



IG 



a ■<- 



characters of a TS 




IG node 
IG internode 



OG node 



OGs 



IG 



OGs 




IG 





a,b 




Fig. 3. 1 . — a. Tree illustrating some general terms used in this chapter, b. Known outgroup relationships. 
c. Unknown outgroup relationships, d. A decisive character polarity decision with "a" at the OG node. e. An 
equivocal character polarity decision with '"a.b" at the OG node. 



CHARACTER ARGUMENTATION AND CODING 27 

derived. The ingroup in Fig. 3.1a is indicated by a polytomy because we presume that the 
relationships among members are unknown. In all other diagrams, the ingroup is indicated 
by a shaded triangle, which means exacUy the same thing as the polytomy but is easier to 
draw. Note that the ingroup always has both characters. The relationships among outgroups 
can either be resolved (Fig. 3.1b) or unresolved (Fig. 3.1c). A decision regarding the 
character found at the outgroup node may be either decisive (Fig. 3. Id) or equivocal (Fig. 
3.1e). If decisive, then we know that the best estimate of the condition found in the ancestor 
of our ingroup is the character in question (in this case, "a" is plesiomorphic and "b" is 
apomorphic). If equivocal, then we are not sure; either "a" or "b" could be plesiomorphic. 

Maddison et al. (1984) treat the problem of polarity as one in which the investigator 
attempts to determine the character to be assigned to the OG node. In effect, it is the 
character of the ancestor of the ingroup and its sister group (first outgroup) that will give us 
information about the characters of the common ancestor of the ingroup. Simple parsimony 
arguments are used in conjunction with an optimization routine developed by Maddison et 
al. ( 1 984) that was built on the earlier routines of Farris ( 1 970) and Fitch ( 197 1 ). There are 
two cases. The first case is relatively complete and is built on known relationships among the 
outgroups relative to the ingroup. The second case is where the relationships among the 
outgroups are either unknown or only partly resolved. Because the first case is the simplest, 
we will use it to describe the general algorithm. 

To illustrate the algorithm, we will use the following character matrix (Table 3.1 ) for the 
hypothetical group M-S. Sidae is the ingroup, and M, N, O, P, Q, and R are outgroups. The 
sister group is PQR. 

Example 3.1. — Character polarity in the group Sidae. 

Table 3. 1 . — Data matrix for the analysis of Sidae (Example 3.1). 











Taxon 








TS 


M 


N 


O 


P 


Q 


R 


Sidae 


1 


b 


a 


a 


b 


b 


a 


a,b 


2 


b 


b 


a 


b 


b 


a 


a,b 


3 


a 


b 


b 


b 


b 


a 


a,b 


4 


a 


a,b 


a 


b 


b 


a 


a,b 



1. Draw the phylogenetic tree of the ingroup and outgroups. You cannot reconstruct the 
entire tree on the basis of the characters in the matrix shown above. These characters relate 
to the resolution of relationships in the ingroup, not to the relationships of the ingroup to the 
outgroup taxa. Presuinably. you have either done an analysis or you have used the analysis 
of another investigator. Figure 3.2 shows the result of this previous analysis with the nodes 
numbered. 



28 



KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 



M N O P Q R Sidae 




Fig. 3.2. — TTie relationships of the Sidae and its closest relatives. Outgroups are letters, nodes are numbers 
(Example 3.1). 



2. For each transformation series, label each of the branches with the character for that 
taxon. Use the character matrix for this task. This has been done for TS 1 in Fig. 3.3. 



M N O P Q R Sidae Taxa 

baa b b a a,b Characters 




OG node 



1 Root node 



Fig. 3.3. — TTie relationships of the Sidae and its relatives, with characters fi"om TS 
relevant nodes labeled (Example 3.1). 



(Table 3.1) and 



3. Proceeding from the most distant branches (in this case, M and N), do the following. 
Label the node "a" if the lower node and adjacent branch are both "a" or "a" and "a,b"; label 
it "b" if the lower node and adjacent branch are "b" or "b" and "a.b." If these branches/nodes 
have different labels, one "a" and the other "b." then label the node "a.b." Note that node 1 is 
not labeled; it is termed the root node. For us to label the root, we would need another 
outgroup. Because we are not interested in the root or in the outgroups, we forget about this 
node. After all, we are supposed to be solving the relationships of the ingroup. Node 2, the 
node immediately above the root, is labeled "a.b" in Fig. 3.4 because the first branch (M) is 
"b" and the second branch (N) is "a." 



CHARACTER ARGUMENTATION AND CODING 



29 



M N 


P Q R Sidae 


Taxa 


baa 


b b a a,b 


Characters 




Fig. 3.4. — First polarity decision for TS 1 , analysis of the Sidae (Example 3.1). 



4. Inspect the tree. Do any of the outgroups have a branching structure? In this example, 
the sister group has a branching structure. For each group of this kind, you need to assign 
values to their lowest node. So, you work down to the lowest node. Assign to the highest 
node in such a group a value derived from its two branches. For example, the value "b" is 
assigned to node 4 in Fig. 3.5. 



M N 


P Q R Sidae 


Taxa 


baa 


b b a a,b 


Characters 




Fig. 3.5. — Second polarity decision forTS 1, analysis of the Sidae (Example 3.1). 



5. Continue in the direction of the ingroup to the next nodes. For this we use a 
combination of previous decisions (labeled nodes) and new information from terminal taxa 
whose ancestral nodes have not been labeled. For example, node 3 in Fig. 3.6 is assigned a 
decisive "a" based on the "a" of O and the "a,b" of node 2. Node 5 is assigned "a,b" based on 
the "a" of R and the "b" of node 4. 



30 



KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 



M N O 
baa 



P Q R Sidae 
b b a a,b 



Taxa 
Characters 




Fig. 3.6. — Third and fourth polarity decisions forTS 1, analysis of the Sidae (Example 3.1). 

6. The analysis is over when we reach an assignment concerning the OG node. In this 
example, the assignment to node 6 is a decisive "a" (Fig. 3.7). 

M N O P Q R Sidae Taxa 

baa b b a a,b Characters 




Fig. 3.7. — Assignment of polarity to the OG node forTS 1. analysis of the Sidae (Example 3.1). 

Figure 3.8 shows characters of TS 2 of the matrix worked out for each node. Note that in 
this case the decision is equivocal for the OG node. 



M N O P Q R Sidae Taxa 

b b a b b a a,b Characters 




Fig. 3.8. — Polarity decisions forTS 2, analysis of the Sidae (Example 3.1). 



CHARACTER ARGUMENTATION AND CODING 31 

One last thing. Each of these decisions is made using a single transformation series at a 
time. This does not mean that equivocal decisions based on single characters taken will 
remain equivocal at the end of the analysis. The final disposition of character states is subject 
to overall parsimony nales. 

Rules of Thumb 

Maddison et al. (1984) present two rules of analysis that can be used when sister group 
relationships are known. These rules will help you bypass some of the argumentation for 
each node of the tree. 

Rule 1: The Doublet Rule. — If the sister group and the first two consecutive outgroups 
have the same character, then that chai'acter is decisive for the OG node. Any two 
consecutive outgroups with the same character are called a doublet. 

Rule 2: The Alternating Sister Group Rule. — If characters are alternating down the tree, 
and if the last outgroup has the same character as the sister group, then the character will 
be decisive for the OG node. If the last outgroup has a different character, then the 
character decision will be equivocal. 



Other Situations 

Maddison et al. (1984) also discuss situations in which the relationships among the 
outgroups are either not resolved or only partly resolved. After you have finished this 
workbook, you should review their discussion on these topics. We will only mention two 
important observations. ( 1 ) Whatever the resolution of the outgroup relationships, the sister 
group is always dominant in her influence on the decision. If the sister group is decisive for 
a particular state, e.g., "a," no topology of outgroups farther down the tree can result in a 
decisive "b." (2) If you are faced with no sister group but only an unresolved polytomy 
below the group you are working on, the frequency of a particular character among the 
outgroups in the polytomy has no effect on the decision for the OG node. For example, you 
could have 10 possible sister groups with character "a" and one with character "b." and the 
decision would still be equivocal at the OG node. Thus, common is not the same as 
plesiomorphic, even among outgroups. 



Quick Quiz — Outgroups and Polarities 

1 . Halfway through your phy logenetic study of the saber-toothed cnidaria, your inquiry suffers 
a fate worse than death. The supposed world's expert. Professor Fenitico. publishes an 
arrangement lumping your group with its sister group, placing them both in the same genus. 
How does this affect your analysis'? 

2. What happens if all the members of the ingroup have a character not found in the sister group 
or any other outgroup? 



32 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



Polarity Exercises 

For each of the trees and tables below, determine the state to be assigned to the OG node 
for each of the transformation series. Show the work for each (labeled) node in the outgroup. 
Prepare a matrix of your decisions using the labeled nodes as taxa. State your decision as 
equivocal or decisive. The monophyly of the ingroup and the relationships among the 
ingroup and the outgroups are assumed in each exercise. (No information in the data 
matrices is relevant to solving the tree shown in each exercise.) 

EXERCISE 3.1. — Detennine the character assignment for TS 3 and 4 from Table 3.1 for 
Example 3. 1 . In TS 4, treat the polymorphic character in taxon N exactly like you would 
treat an equivocal decision at a node. 

EXERCISE 3.2.— Use Table 3.2 and the tree in Fig. 3.9. 

Table 3.2. — Data matrix for Exercise 3.2. 









Transformation 


series 






Taxon 


1 


2 


3 


4 


5 


6 


A 


a 


a 


a 


a 


a 


a 


B 


a 


a 


a 


b 


a 


b 


C 


b 


a 


a 


a 


b 


a 


D 


b 


b 


a 


b 


b 


b 


E 


b 


b 


b 


a 


a 


b 


IG 


a,b 


a,b 


a,b 


a,b 


a,b 


a,b 



ABODE IG 




Fig. 3.9. — Tree for Exercise 3.2. 



CHARACTER ARGUMENTATION AND CODING 



33 



EXERCISE 3.3.— Use Table 3.3 and the tree in Fig. 3.10. 



Table 3.3. — Data matrix for Exercise 3.3. 









Transformation series 






Taxon 


1 


2 


3 4 


5 


6 


A 


a 


b 


a a 


a 


a 


B 


a 


a 


b a 


a 


a 


C 


b 


b 


a b 


a 


b 


D 


b 


b 


b a 


b 


a 


E 


b 


b 


a a 


b 


b 


IG 


a,b 


a,b 


a,b a,b 


a,b 


a,b 



B C D E 



IG 




Fig. 3.10. — Tree for Exercise 3.3. 



EXERCISE 3.4.— Use Table 3.4 and the tree in Fig. 3.11. 



Table 3.4. — Data matrix for Exercise 3.4. 









Transformation 


series 






Taxon 


1 


2 


3 


4 


5 


6 


A 


a 


a 


a 


a 


a 


a 


B 


a- 


b 


b 


b 


b 


b 


C 


a 


b 


a 


b 


b 


b 


D 


b 


b 


a 


b 


b 


b 


E 


b 


b 


b 


a 


a 


b 


F 


a 


b 


a 


b 


a 


a 


G 


a 


a 


a 


a 


a 


a 


IG 


a,b 


a,b 


a,b 


a,b 


a,b 


a,b 



34 



KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 



B C D E F G 



IG 




Fig. 3. 1 1 . — Tree for Exercise 3.4. 



EXERCISE 3.5.— Use Table 3.5 and the tree in Fig. 3.12. 



Table 3.5. — Data matrix for Exercise 3.5. 









Transformation series 






Taxon 


1 


2 


3 4 


5 


6 


A 


a 


a 


a a 


a 


a 


B 


a 


b 


a b 


a 


a 


C 


b 


a 


a a 


b 


b 


D 


b 


a 


a b 


b 


b 


E 


a 


a 


a b 


b 


b 


F 


a 


a 


b b 


b 


b 


G 


b 


b 


b b 


b 


b 


H 


b 


b 


b a 


b 


b 


I 


b 


b 


a b 


b 


b 


J 


a 


b 


a a 


a 


b 


K 


a 


b 


a b 


a 


a 


IG 


a,b 


a,b 


a,b a,b 


a,b 


a,b 



Character Coding 



As you learned in Chapter 1 , a chaiacter is a feature of an organism. A character code is a 
numerical or alphabetical symbol that represents a particular character. We have already 
used codes in our previous exercises. By using these characters and their codes, you have 
learned something about the basics of tree reconstruction using classical Hennig 
argumentation and some of the approaches to detemiining the polarity of characters through 
character argumentation. You already know something about different kinds of characters, 
homologies, analogies, and homoplasies. hi this section, you will be introduced to some of 



CHARACTER ARGUMENTATION AND CODING 



35 



B 



D 



H 



K 



IG 




Fig. 3.12. — Tree for Exercise 3.5. 



the different kinds of derived characters encountered in phylogenetic research and some of 
the problems associated with assigning codes to these characters. Before you begin, it might 
be useful to reread the sections in Chapter 1 about the attributes of specimens. (You should 
be aware that some investigators refer to transformation series as "characters" and char- 
acters as "character states." It is usually quite cleai- what is being discussed, but this is a 
potential source of confusion.) 

All of the derived characters we have dealt with up to this point are 1 ) qualitative 
characters and 2) part of binary transfomiation series. A binary transfoiTnation series 
consists of a plesiomorphy and its single derived homologue. By convention, the 
plesiomorphy is coded "0" and the derived homologue is coded " 1 ." As we mentioned in 
Chapter 1 , such binary transformation series are already ordered by virtue of the fact that 
they are binary. When an investigator works on a large group, or even a small group that has 
undergone considerable evolution, she may find that there ai'e several different homologous 
characters in a transformation series. For example, if she were researching the phylogenetic 
relationships of fossil and Recent horses, the transfomiation series containing the characters 
for the number of toes of the hind foot would contain four different but related characters: 
four toes, three toes, and one toe in the ingroup and five toes in the outgroups. This kind of 
transformation series is termed a multistate transformation series. A multistate 
transformation series contains a plesiomoiphic character and two or more apomorphic 
characters. 

Simple binary transformation series present no problem in coding. The investigator codes 
by outgroup argumentation, according to the information available, producing a matrix full 
of and 1 values. You have practiced this kind of coding in Chapter 2. Complications arise 
if there are one or more polymorphic taxa, i.e., taxa with both the plesiomorphic and 



36 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

apomorphic characters. The problem is only critical when both types of characters are found 
in a single species. In such cases, the taxon can be treated as having both characters, a 
designation easily handled by available computer prograins. 

Multistate transformation series can be grouped in two ways: 1 ) according to what we 
know of their evolution and 2) according to the way they are related. Ordered 
transformation series are those in which the relationship of characters within the 
transformation series are specified (and presumably "known"). Binary transformation series 
are the simplest case of ordered transformation series. A polarized ordered transformation 
series is one in which not only the relationships are specified, but also the direction of 
evolution. Unordered transformation series are those where the relationships of characters 
to one another are not specified. This does not mean that we know nothing about the 
transformation series. Frequently, we know which of the characters is plesiomorphic; we 
just don't know the order of the derived transformation series. Such a transformation series 
is partly polarized. If you are using a computer program such as PAUP or MacClade, the 
program can tell which of the unordered characters is the plesiomorphic one when you root 
the tree at a specific point or with a specific taxon. The program relies on you, however, to 
specify this information correctly. 

The second way of grouping multistate transformation series is by their relationships. 
Polarized transformation series may come in several varieties. In the simplest case, the 
characters might be related in a linear fashion. A linear transformation series consists of 
characters related to one another in a straight-line fashion such that there are no branches on 
the character tree (Fig. 3.13). The relationships of these characters can be termed a 
character tree. It is important to understand that a character phylogeny is not the same as a 
phylogeny of taxa. A character tree contains only information about the relationships among 
characters; the distribution of these characters among taxa is shown in descriptions, 
diagnoses, and character matrices. 

OG, A B, C, D E, F G, H 

Fig. 3.13 . — A simple linear character tree of four characters. Letters represent taxa in which each character 
is found. 



Linear transformation series present no problems in coding; one simply assigns a value to 
each character in ascending order. Each value is placed in the data matrix in a single column, 
and each apomorphy contributes to the length of the tree in an additive fashion. We use the 
term additive because each instance of evolution is one step along the tree, and counting all 
of the steps in a straight line shows exactly how much the transformation series has added to 
the overall tree length. (Such transformation series are often termed additive multistate 
characters.) 



CHARACTER ARGUMENTATION AND CODING 



37 



A branching transformation series contains characters that are not related to each other 
in a straight-hne fashion (Fig. 3.14). Such transformation series may present problems 
because the relationships among the characters are represented by a branching pattern rather 
than a straight-line pattem. Because of this, the characters cannot be coded in an additive 
fashion. Such transformation series are also called nonadditive or complex transformation 
series. Because the characters are not related in a linear fashion, simple additive coding will 
result in errors in translating the transformation series into a phylogenetic tree. 



H 



E, G 



B, D 



^\ OG, A 



Fig. 3.14. — A complex branching character tree of six characters. Letters represent taxa in which each 
character is found. 



We present two examples of character coding using three techniques. You will probably 
ask where we came up with the character trees of the two transformation series. This is a 
good question and one we will return to in a later section. For now, we are only concerned 
with the formalities of coding and not how one actually determines the character trees. 

Example 3.2. — A simple linear transformation series. 

hi Fig. 3.13, we show a simple linear transfonnation series of four characters. Below the 
character tree is an account of the distribution of these characters among nine taxa. The data 
matrix for this transformation series can be constructed in one of two basic ways. First, we 
can simply code the transformation series in a linear fashion, assigning a value to each 
character based on its place in the character phylogeny. We have chosen to code with values 
ranging from to 3 (Table 3.6). 



38 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

Table 3.6. — Data matrix for a simple linear transformation series coded by the linear and the additive 
binary methods (Example 3.2). 



Additive binary coding* 



Taxon 



coding 


c + c/c + 


C/D 


C/C 


: + c/D 


C/D 


































1 














1 














1 














2 








1 





2 








1 





3 








1 


1 


3 








1 


1 



OG 

A 

B 

C 

D 

E 

F 

G 

H 



* Column heads are apomorphic ( 1 ) characters. C = circle; C/C = circle/circle; C/D = circle/dot. 

We could also use additive binary coding, which is a method that breaks the character 
down into a number of subcharacters, each represented by its own column of information. 
For example, because the characters in the transfomiation series really consist of subsets of 
related characters, we can consider both "circle/circle" and "circle/dot" as subsets of "circle" 
because each is derived from "circle" later in the character tree. The first additive binary 
column in Table 3.6 reflects this fact, coding "oval" as the plesiomorphic character (0) and 
"circle" plus all of its descendants as apomorphic ( 1 ). "Circle/dot" is a subset of "circle/ 
circle." Both "oval" and "circle" are plesiomorphic relative to "circle/circle" so they get a 
coding of "0," whereas "circle/circle" and its descendant "circle/dot" get a coding of "1." 
Finally, "circle/dot" is apomorphic relative to "circle/circle" (and "oval" and "circle") so it is 
coded "1" in the third coluiTin. So in total, we have produced three columns to represent the 
transformation series. Now, go along the rows and add up all of the 1 "s in the additive 
binary matrix and put them into a single column. You should find that you have replicated 
the original linear transformation series of 0-1-2-3. Either method of coding produces 
exactly the same phylogeny. But, there are some differences. If you use binary coding, you 
must keep in mind that formal computer algorithms and the programs that use them cannot 
tell the difference between three noncorrelated and independent transformation series and a 
single binary coded transfonnation seiies. This doesn't cause problems with phylogenetic 
analysis, but it can produce results that seem strange in biogeographic analysis and the 
analysis of coevolutionary patterns, which we will discuss in Chapter 7. 

Example 3.3. — A branching transfonnation series. 

Let us look at the branching transformation series in Fig. 3.14. The taxa sharing a 
particular character are shown beside or above the character on the character tree. This 
transformation series is considerably more complex than the first one. It should be obvious 
that a single labeling of characters in a linear fashion would result in some misinformation. 
How do we show these complex relationships? There are two basic methods, nonadditive 



CHARACTER ARGUMENTATION AND CODING 39 

binary coding and mixed coding. Because we have already seen an example of binary 
coding, let us tum to this method first. 

Review the character tree and then examine the nonadditive binary codings in Table 3.7. 
Note that "square" is apomorphic relative to "triangle," by outgroup comparison, and that 
"square" is ancestral to all other characters in the character tree. Our first binary column 
reflects this fact: "square" and all of its descendants are coded " 1 ." whereas "triangle" is 
coded "0." "Square/square" is derived from "square." "Rectangle" is also directly derived 
from "square." Look at "square/square." It is only found in taxon C. We produce a new 
column reflecting this fact, hi this column "square/square" acts like an autapomorphy 
(which it is). "Rectangle" does not act as an autapomoiphy; it is plesiomorphic to two other 
characters. This fact is used to code "rectangle" in a similar manner to the way we coded 
"square," as shown in the third column. Since both "rectangle/triangle" and "rectangle/dot" 
are unique to their respective taxa, F and H, we code them in a manner similar to "square/ 
square." Now, you are able to reconstmct the phylogeny of the group AH using the two 
character phylogenies. 

Table 3.7. — Data matrix for a branching transformation series coded by the nonadditive binary and the 
mixed methods (Example 3.3). 







Nonadditive 


binary 


coding* 






Mixed 


coding* 




Taxon 


All except T 


S/S 


R+ 


R/D 


R/r 


T+S+R+R/D 


S/S 


R/r 


OG 




























A 




























B 
















1 










C 




1 











1 




1 





D 
















1 










E 







1 








2 










F 







1 


1 





2 







1 


G 







1 








2 










H 







1 





1 


3 











* Column heads are apomorphic (>1 ) characters. T = triangle; S = square; S/S = square/square; R+ = rectangle and 
all descendants; R/D = rectangle/dot; R/T = rectangle/triangle. 

Mixed coding is a hybrid between additive binary coding and linear coding. Mixed 
coding has also been termed nonredundant linear coding. By convention, the longest 
straight-line branch of the character tree is coded in a linear fashion. Branches off this linear 
tree are coded in an additive binaiy fashion. This strategy might save character columns, 
depending on the asymmetry of the character tree. We can code the section of the character 
tree that goes "triangle," "square," "rectangle," "rectangle/dot" in a single column (0-1-2-3) 
in the first column of Mixed coding in Table 3.7. (How do you know to use "rectangle/dot" 
as the fourth character in the transfonnation series? Actually, the choice is completely 
arbitrary; remember, nodes can be freely rotated. We could have just as well used "rectangle/ 
triangle" and coded "rectangle/dot" as the autapomorphy.) A separate column is then used 
for "square/square" (column 2), and a final column for "rectangle/triangle." 



40 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

Basal bifurcations occur when both the outgroup and one of the ingroup taxa have the 
same character. In this case, assign "1" rather than "0" to the most plesiomorphic character 
and proceed. This coding strategy serves to "link" the columns and will not add steps to the 
tree. 

Note on character coding. Newer computer algorithms such as PAUP 3.0 can use a 
character tree directly if the investigator inputs the relationships of the characters. It then 
uses this information to construct an additive binary matrix for analysis, which the 
investigator never sees (Swofford, 1990). 



Quick Quiz — Character Coding 

1 . Your research into the systematics of the spade-lipped mugmorts is halted until you resolve 
the coding of a troublesome transformation series. You have identified the plesiomorphic 
character, but the remaining six cannot be polarized. What type of transformation series 
should be considered? 

2. The copulatory organ of spade-lipped mugmorts has various colors, including no color at all. 
The evolutionary sequence of color change is not known except that the sister group and all 
other outgroups have colorless organs. Although you could opt for a polarized but unordered 
transformation series, you opt instead for a binary transformation series for each color (e.g., 
no color [0] to blue [ 1 ], no color [0] to green [1], etc.). What effect will your decision have on 
reconstructing the tree? 



Coding Exercises 

For each of the trees shown below do the following: 

1 . Determine the possible types of coding strategies that might be used and list them. 

2. Explain why certain coding strategies cannot be employed for the particular character tree. 

3. Prepare a data matrix for each type of coding strategy you think could be employed. 

4. Solve the phylogenetic problem with the data in the matrix. 

EXERCISE 3.6.— Use Fig. 3.15. 

I >- m >■ n >■ 

OG A B C 

Fig. 3.15. — A character tree for four characters. Capital letters represent taxa in which each character 
(lowercase letters) is found. We use letters rather than numbers to emphasize the difference between a 
character and a character coding. 



CHARACTER ARGUMENTATION AND CODING 



41 



EXERCISE 3.7.— Use the character tree in Fig. 3.16. 




Fig. 3.16. — A character tree for eight characters. Capital letters represent taxa in which each character 
(lowercase letters) is found. 



EXERCISE 3.8.— Use the character tree in Fig. 3.17. 




m OG 



Fig. 3.17. — A character tree for nine characters. Capital letters represent taxa in which each character 
(lowercase letters) is found. 



42 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

Chapter Notes and References 

1. Discussions of homology, different kinds of characters, and basic character 
argumentation can be found in Wiley (1981a). However, this reference is outdated when it 
comes to outgroup comparison and contains no information of use on character coding and 
other "modem" issues. 

2. There is a lot of information on outgroups and outgroup comparisons. Maddison et al. 
(1984) was preceded by Watrous and Wheeler (1981) and a criticism of them by Farris 
(1982). Wiley (1987b) contains a summary of the three papers. The discussion by Crisci and 
Stuessy (1980) is, in our opinion, positively misleading and should be avoided. Donoghue 
and Cantino (1980) discuss one method of outgroup comparison, the outgroup substitution 
method, that can be useful when relationships among outgroups are problematic. 

3. Those who have read some phylogenetic literature will note that we have avoided, 
until now, any mention of other criteria. Wiley ( 1 98 1 a) discusses several other criteria. Other 
useful discussions can be found in de Jong (1980) and Stevens (1980). The major bone of 
contention is what is known as the ontogenetic criterion. Some, such as Nelson (1978, 
1985), Patterson (1982), Rieppel (1985), and Weston (1988), advocate the ontogenetic 
criterion as a (or the) major criterion for detemiining polarity. We do not think this is a 
general criterion (see Brooks and Wiley, 1985; Kluge, 1985, 1988a; O'Grady, 1985; Kluge 
and Strauss, 1986), but we recognize that it can be used to both check hypotheses of 
homology (cf. Hennig, 1966; Wiley, 1981a; Patterson, 1982; Kluge. 1988a) and infer 
polarity under certain assumptions. Before you employ this criterion, you should read 
Mabee(1989). 

4. Some papers of interest on character coding include Farris et al. (1970), Mickevich 
(1982), O'Grady and Deets (1987), Pimentel and Riggins ( 1987), and O'Grady et al. (1989). 

Complications arise if one or more of the taxa have both the plesiomorphic and 
apomorphic character. Polymorphic taxa have both the plesiomoiphic and apomorphic 
characters. Actually, the problem is only critical when both are found in a single species. 
Considerable controversy surrounds the coding of such characters, especially when 
biochemical characters are used. There are two ways of handling such characters: 1 ) coding 
the taxon as having the apomorphy only and discounting the plesiomorphy or coding both 
characters as present and using a computer program such as PAUP that can handle 
polymorphic data cells (qualitative coding) or. 2) coding according to frequency of each 
character. Swofford and Berlocher (1987) present a strong case for analysis of frequency 
data and suggest computational methods for accomplishing this within a phylogenetic 
analysis. D. L. Swofford (pers. comm.) has authored a computer program (FREQPARS) to 
accomplish this task. Buth (1984) is an excellent introduction to the use of electrophoretic 
characters. 



CHARACTER ARGUMENTATION AND CODING 43 

Quick Quiz Answers 
Outgroups and Polarities 

1. This question has no simple answer. We suggest the following. Examine the paper. Has Professor 
Fenitico provided synapomorphies to support his argument? If not, then he has only produced another 
arrangement and not a scientific hypothesis you can evaluate, so you should proceed with your problem as if 
nothing had been published. If he does provide synapomorphies, what is the nature of these characters? Do 
they demonstrate that the sister group is monophyletic? Is the sister group still the sister group, even if it is 
now in the same genus? If so, then the nature of our character argumentation has not changed, only the 
taxonomy, which might be very important to Professor Fenitico but should not be important to you. However, 
if Professor Fenitico has demonstrated that the supposed sister group is really embedded within your group, 
then take this into consideration, redesign your arguments, and write Professor Fenitico to tell him that names 
really don't mean anything, especially his. 

2. If this character is really unique to the ingroup, then it is a synapomorphy of the members of the group 
(or, if you wish, an autapomorphy of the group). 

Character Coding 

1 . You can opt for an unordered transformation series or you can try coding six binary ordered series. If 
you pick the binary series, check answer 2. 

2. Because "no color" is symplesiomorphic, repeated use of this character in different transformation 
series will result in an answer that has no bearing on the relationships among the taxa. The effect is to render 
all of the color characters autapomorphic. which implies that all are independently derived from "no color." 
Better see answer I and opt for an unordered transformation series. 



44 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 1 9 



Chapter 4 

TREE BUILDING AND OPTIMIZATION 



Phylogeneticists frequently describe their work in terms of "building trees" or 
"reconstructing phylogenies." These activities are directed towards attempts to discover 
something we believe exists in nature, the common ancestry relationships among the 
organisms we study. Interestingly, modem computer programs do not spend much 
computing time building trees. Rather, most of the time is spent evaluating different tree 
topologies (branching patterns) in an effort to find the tree that meets a criterion of optimality 
given the data. How the tree is actually generated may be irrelevant. For example, you can 
evaluate all of the possible trees for a three taxon problem by simply mapping the character 
distributions on the four possible trees in the most efficient manner (i.e., maximizing the 
number of synapomorphies and minimizing the number of homoplasies needed given the 
tree). You don't have to build a tree, all the possible trees are given. Under the criterion that 
the shortest tree is the optimal tree, all you have to do is count the changes and pick the 
shortest tree among the four possibilities. 

Of course, as the number of taxa increases, the number of possible trees increases very 
quickly (see Chapter 6). Because phylogenetic methods were originally built around 
constructing trees, many of the classic works emphasize reconstruction, and they do so using 
methods such as Hennig argumentation (Hennig, 1966) and the Wagner algorithm (Kluge 
and Farris, 1969). Although many "modem" phylogeneticists will never use classic Hennig 
argumentation and algorithms such as the Wagner algorithm provide only the starting point 
for some (not even all) modem computer programs, it is important for you to get the feel of 
these approaches because they give insight into the nature of phylogenetic data and help you 
understand how previous investigators arrived at their conclusions. Thus, we have organized 
this chapter in a quasi-historical fashion. We begin with Hennig argumentation, Hennig 's 
own method for reconstructing phylogenies. We then use the Wagner algorithm to teach the 
rudiments of a formal algorithm and how such an algorithm might be implemented and 
provide some basic terms encountered in more modem methods. We then discuss the 
concepts of the optimal tree, optimal character distribution, various parsimony criteria, and 
ACCTRAN and DELTRAN optimization. Finally, we provide a very brief discussion of 
how current algorithms operate to produce optimal or near optimal trees. 

Hennig Argumentation 

You already have had practice at performing analyses using Hennig argumentation in 
Chapter 2. However, you did it in a rather laborious way, using the inclusion/exclusion 
principle. Hennig argumentation was the original phylogenetic algorithm, and its application 
is still common. For simple problems, Hennig argumentation presents no technical 



46 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



problems. However, even with relatively few taxa and characters, you will find it much too 
tedious to make all of those single-character trees and all of the logically incompatible 
alternative trees. In computer implementations of phylogenetics, the computer perfomis this 
boring task. An investigator working without a computer has two problems. Fii'st, she would 
never want to make all the inclusion/exclusion trees. Second, with even a small amount of 
homoplasy the investigator runs the chance of missing equally parsimonious solutions to the 
tree that she finds, not because the algorithm is defective but because the human mind rejects 
the ennui of considering all possible alternative trees. A really experienced phylogeneticist 
instead will "inspect" the data matrix and produce a first tree based on this general 
inspection, filtering through the data in her mind. We would like you to try this on the 
exercises below. One strategy, for example, would be to circle potential groupings within the 
data matrix. Another strategy is to start with "obvious groups" (=lots of synapomorphies) 
and then attempt to link them together. Although this might seem rather imprecise to you, 
remember that it is similar to the method Hennig himself probably used. Remember, no tree 
you draw has to be the final tree. All trees are hypotheses. 

Hennig Exercises 

EXERCISE 4.1.— Bremer's (1978) Leysera data. 

Leysera is a small genus of composite shrublets found in southern Africa (three species) 
and the Mediterranean region (one species). The closest relatives of Leysera appear to be the 
genera /l//i77.\7a, Rosenia, and Relhandia. The phylogenetic relationships of these genera are 
as follows. Leysera, Rosenia. and Relhandia fomi a trichotomy. Athrixia is the sister of these 
three genera. Leysera is monophyletic based on two characters: 1 ) the chromosome number 
is 2N = 8 and 2) all have a solitary capitula on a long peduncle. The distribution of characters 
among the four species oi Leysera is given in Table 4. 1 . 

Table 4. 1. — Leysera ingroup characters. 

Floret Pappus Achene Pappus Life 

Taxon Receptacle tubules type surface scales cycle 



L. longipes 


smooth 


with glands 


barbell ate 


smooth 


subulate 


perennial 


L. leyseroides 


rough 


with hairs 


plumose 


rough 


wide, flat 


annual 


L. tennella 


rough 


with hairs 


plumose 


rough 


wide, flat 


annual 


L. gnaphalodes 


rough 


with hairs 


plumose 


rough 


subulate 


perennial 



Based on outgroup information, the following characters are plesiomorphic: 1) recep- 
tacles smooth, 2) hairs absent on the floret tubules, 3) barbellate pappus, 4) achene surface 
smooth, 5) pappus scales subulate, and 6) perennial life cycle. 



TREE BUILDING AND OPTIMIZATION 



47 



1. Prepare a data matrix. 

2. Analyze the phylogenetic relationships of Leysera based on the information given and 
draw the tree of relationships. 

EXERCISE 4.2.— Siegel-Causey's (1988) cliff shags. 

Shags, cormorants, and anhingas comprise a clade of marine and littoral fish-eating birds. 
Among shags, the genus Stictocarho (cliff shags) comprises eight species. In this exercise, 
you will use both outgroup infomiation and ingroup infonnation to reconstruct the 
relationships among six species of Stictocarbo. The seventh species (S. magellanicus) is 
used as the sister group, and species in other genera provide additional outgroups. 

1 . Using the tree in Fig. 4. 1 and the characters in Table 4.2, determine the character at the 
outgroup node for each transfomiarion series and arrange this as a character vector labled "OG." 



//. 



.\o 



.# 



^9 



r / 



# 



.<>' 




.^^^ 



Fig. 4.1. — A hypothesis of the phylogenetic relationships among certain shags. 

2. Adding the characters in Table 4.3, reconstruct the phylogenetic relationships among 
the remaining six species of the genus. Tip: Some of the decisions you reached in step 1 are 
equivocal. Don't use these transformation series to reconstruct the initial tree. 

3. After reconstructing the relationships among the species, examine the transformation 
series with equivocal decisions at the outgroup node. Can you now characterize them? Will 
one or more remain equivocal, giving rise to alternative inteipretations of character 
evolution? 



The Wagner Algorithm 

The basics of the Wagner algorithm were published by Kluge and Farris (1969). Although 
phylogenetic, the algorithm was developed independent of the Hennig argumentation 
algorithm and was based on Wagner Groundplan Divergence Analysis. 



48 



KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 



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TREE BUILDING AND OPTIMIZATION 49 

Wagner Definitions 

Like all techniques, the Wagner algorithm comes with certain terms that must be leamed 
to appreciate its logic. In some cases, these terms are illustrated by examples (refer to Table 4.4). 

Table 4.4. — Sample data matrix for Wagner definitions and Exercise 4.3. 



Taxon 






Characters 






M 


1 














A 





1 





1 


1 


B 





1 


1 









1 . A particular character (X) of a particular taxon (A) is defined as X(A./'). where / is the 
/th character in a vector of / characters. 

2. The vector of characters for a particular taxon is defined as X X(A,/). For example, the 
character vector for M is 

I X(M,/) =10 0. 

3. The difference (D) between two taxa is the sum of the absolute differences between 
their characters: 

D(A,M) = I IX(A,0 - X(M,/)I. 

We calculate this in the following manner: 

D(A,M) = I IX( A,/) - X(M,/")I 

= 10-11 + 11-01 + 10-01 + 11-01 + 11-01 
= 4. 

EXERCISE 4.3. — Calculate the difference (D) between A and B and between M and B 
(Table 4.4). 

4. The interval (ENT) of a taxon is the length of the line between that taxon and its 
ancestor. For example, the interval of B is 

INT(B) = D[B,ANC(B)], 

where INT(B) is the interval of taxon B, ANC(B) is the hypothetical ancestor of B, and 
D[B,ANC(B)] is the path length distance of B to its ancestor. 

Example 4.1. — Calculating interval B. 

Let us take our simple data matrix and calculate an interval. Designate M as the ancestor 
(it's really the outgroup, hut it doesn't matter here). So. ANC(B) is M and the formula reads 



50 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



INT(B) = D[B.M] 

= IIX(B./)-X(M./)l 

= 10-11 + 11-01 + 11-01 + 10-01 + 10-01 

= 3. 



The interval is shown graphically in Fig. 4.2. 




INT(B) = 3 



Fig. 4.2. — Graphic representation of INT(B) of the Wagner algorithm. 



EXERCISE 4.4.— Distances and intervals (Table 4.5). 



Table 4.5. — Data matrix for Wagner algorithm distances and intervals (Exercise 4.4). 



Taxon 








Characters 








ANC 






















A 


1 


1 
















B 


1 





1 







1 





C 


1 





1 




1 





1 



For this exercise we will make the following assumptions. First, all of the characters 
scored "1" have been previously argued as derived, using the standard outgroup procedures. 
Second, we have used outgroup comparison to construct a valid hypothetical ancestor. (Note 
that the ancestral vector is composed entirely of "0" values.) From the data in Table 4.5, 1 ) 
calculate the distance of each taxon to each other taxon and from each taxon to the ancestor 
and 2) pick the least distance and calculate the interval from that taxon to the ancestor. 



The Algorithm 

Given any matrix of characters, we implement the Wagner algorithm in the following 
manner (from Kluge and Farris, 1969). 

1 . Specify an ancestor or outgroup. 

2. Within the ingroup, find the taxon that shows the least amount of difference from the 
ancestor/outgroup. To accomplish this, calculate D for each taxon to the ancestor/outgroup. 

3. Create an interval for the taxon that has the smallest D. 



TREE BUILDING AND OPTIMIZATION 5 1 

4. Find the next taxon that has the next smallest difference from the ancestor/sister group. 
Do this by inspecting the original D values you calculated. If there is a tie (i.e.. two or more 
with the same value of D), then arbitrarily pick one. 

5. Find the interval that has the smallest difference with the taxon selected in step 4. To 
compute D(taxon,interval), use the following formula: 

D[B INT{A)] = D(B,A) + D[B,ANC(A)] - D[A,ANC(A)] 

2 

6. Attach the taxon to the interval you have selected by constructing a hypothetical 
ancestor for the two taxa. The character vector of the ancestor, and thus its position along the 
interval, is computed by taking the median value of the existing taxon, its ancestor, and the 
added taxon. 

7. Go to step 4 and repeat for each remaining taxon. 

Example 4.2. — First Wagner calculations. 

We begin with an extremely simple example taken from Wiley (1981a). There are three 
taxa and their ancestor. Step 1 is specified in Table 4.6 as the ANC vector. 

Table 4.6. — Data matrix for Wagner calculations (Example 4.2). 



Taxon 








Characters 








ANC 






















A 


1 


1 
















B 


1 





1 







1 





C 


1 





1 




1 





1 



2. Calculate D from the ancestor for each taxon. You already did this in Exercise 4.5. 

D(A,ANC) = 2 
D(B,ANC) = 3 
D(C,ANC) = 4 

3. Construct an interval for this taxon. A has the smallest distance to ANC, so we 
construct INT(A,ANC): 

INT(A,ANC) = D(A,ANC) = 2. 

4. Select the next taxon that has the smallest D to the ANC. This would be taxon B. 

5. Find the interval that has the smallest D to taxon B. Because there is only one interval, 
INT(A,ANC), we have no choice but to add B to this interval. Therefore, we don't have to 
compute D[B,INT(A)]. We connect B to INT(A) by constructing a hypothetical ancestor 
(X) whose characters are the median of the transformation series of ANC, A, and B, the three 
taxa involved in the problem at this point (Table 4.7). Our tree now has a branch, a new 



52 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



hypothetical ancestor, and, most importantly, three intervals, INT(A), INT(B), and INT(X) 
(Fig. 4.3). 

Table 4.7. — Data matrix for Wagner calculations with a hypothetical ancestor (X) (Example 4.2). 



Taxon 








Characters 








ANC 






















A 


1 


1 
















B 


1 





1 







1 





X (median) 


1 




















A (1 1 000) 



INT(A) 




B(1 1 1 0) 



INT(B) 



INT(X) 
ANC (000000) 
Fig. 4.3. — A branched Wagner tree with a new hypothetical ancestor (X) and three intervals (Example 4.2). 



6. We return to step 4 of the algorithm, adding the next taxon that shows the least 
difference from ANC. By default, this is taxon C, but where do we add C? The algorithm 
states that it should be added to that interval that has the smallest difference from C. 
Therefore, we must calculate three interval difference values, one for each interval in the 
tree. The formula for figuring the difference between a taxon and an interval requires finding 
the difference between the taxon added and the ancestor of the taxon already in the tree, hi 
this case, there are two ancestors. You already know the difference between C and ANC (D 
= 4, see above). But, you haven't calculated the difference between A, B, or C and the new 
ancestor, X, and you haven't calculated the differences between C and A or C and B. This is 
the first step. 

D(A,X) = IX(A./) - X(X,/)I = 1 

D(B,X) = IX(B,/)-X(X,/)l = 2 

D(C,X) = IX(C,/) - X(X,/)I = 3 

D(C,A) = IX(C,/) - X(A,/)I = 4 

D(C,B) = IX(C,/)-X(B,/)l = 3 
D(X,ANC) = IX(C,/) - X(ANC,/)I = 1 



TREE BUILDING AND OPTIMIZATION 



53 



Now we can begin our calculations. 

D(C.A) + D(C,X)-D(A.X) 



D[C,INT(A)] = 



(4 + 3-1) 



= 3 



D[C.INT(B)] = 



D(C,B) + D(C,X)-D(B,X) 



(3 + 3-2) 



= 2 



D[C,INT(X)] = 



D(C,X) + D(C,ANC) - D(X,ANC) 



(3 + 4-1) 

2 
= 3 

Because the difference between C and INT(B) has the smallest value, we construct 
another hypothetical ancestor (Y) and connect C to the tree through this new ancestor to 
INT(B) (Fig. 4.4). To calculate the character vector for this new ancestor, take the median of 
the vectors of the three appropriate taxa, X, B, and C (Table 4.8). You have now completed 
the problem (Fig. 4.4). 



C(1 1 1 1) 

INT(C) 



B(1 1 000) 




ANC 
Fig. 4.4. — Complete Wagner tree with two hypothetical ancestors (X and Y) (Example 4.2). 



54 KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 

Table 4.8. — Complete data matrix for Wagner calculations with a second hypothetical ancestor (Y) 
(Example 4.2). 



Taxon 








Characters 








X 


1 



















B 


1 





1 







1 





C 


1 





1 




1 





1 


Y (median) 


1 





1 














Wagner Tree Exercises 

The two exercises use the same data matrices as those used in the Hennig argumentation 
exercises. To follow the exercise answers more closely, we suggest that you label the 
ancestors sequentially beginning with "A" and follow the tips in each exercise. 

EXERCISE 4.5.— Bremer's (1978) Leysera data. 

Reconstruct the relationships among Bremer's species (Table 4.1) using the Wagner 
algorithm. Use a zero vector ancestor. Tip: There is a tie; add L. leysewides before L. 
tennella if you want to conform directly with the answers. 

EXERCISE 4.6.— Siegel-Causey's (1988) cliff shags. 

Reconstruct the relationships of Siegel-Causey's cliff shags using Table 4.3 and the 
character vector you reconstructed at the OG node. Some of the character decisions at the 
OG node were equivocal; for this exercise use the following decisions for these transfonna- 
tion series: 2-0, 36-1, 102-0, 124-1. This is a large and complicated exercise, but it will be 
worth the effort to complete. Tips: 1 ) To decrease the work load, compare ancestral intervals 
before you compare terminal intervals. If the distance to the interval of, say, ancestor B is 
greater than that to ancestor C, then you will not have to compute the distances to those 
terminal intervals connecting to ancestor B. This shortcut works because we are using path 
length distances. 2) You will find two ties: add 5. aristotelis before 5. wile and add S. 
gaimardi before S . featherstoni if you want to directly follow the answers. Of course, this 
addition sequence is arbitrary. 

Optimal Trees And Parsimony Criteria 

An optimal tree is one of many possible trees that meets a particular criterion. Although 
several criteria exist (i.e., maximum likelihood, parsimony, least square methods for dis- 
tance data; see Swofford and Olsen [1990] for a readable review), we are concerned only 
with parsimony as reflected by measurements such as tree length. By this criterion, usually 
referred to as the maximum parsimony criterion, the optimal tree is the tree that has the 
shortest length for a particular data set given our assumptions about character evolution. 



TREE BUILDING AND OPTIMIZATION 55 

There may be only a single optimal tree, but there may also be several to many optimal trees. 
We usually refer to these as the set of equally parsimonious trees. 

We can consider the distribution of a particular transfonnation series as "optimal" for a 
particular tree topology under the maximum parsimony criterion if that distribution provides 
a narrative explanation of character evolution that minimizes the number of homoplasies 
and maximizes the number of apomorphies. Homoplasy is not precluded, only minimized 
for a particular tree topology. Carefully note that we have referred to character optimization 
given a particular tree topology. Optimization is an a posteriori activity; it does not help 
build trees but is used for evaluating trees that are already built. 

The assumptions about character evolution are reflected by several factors, including how 
we coded the characters (Chapter 3) and how we weight the transformations that have 
occurred. The weight of a transfomiation from one character to another represents an 
assumption on the part of the investigator about the nature of character evolution, which is 
reflected in the kind of parsimony she picks (actually it's more complicated than this, but this 
chaiacterization will do for a start). Up to now. we have assumed that all characters have 
equal weight, which amounts to assigning an equal "cost" of evolution for one step for each 
transfomiation within a transformation series. However, you can imagine that the weight of 
a transformation can be larger or smaller than one. 

Swofford and Olsen (1990) review four criteria that provide an introduction to different 
views on the nature of parsimony. 

1. Wagner parsimony (Kluge and Farris, 1969; Farris, 1970) treats characters as 
ordered such that the change from one character to another implies change through any 
intervening characters in the transformation series. Characters are allowed to reverse freely. 

2. Fitch parsimony (Fitch, 1971) treats characters in a transformation series as 
unordered such that change from one character to another character within a transformation 
series does not imply changes through possible intervening characters. Characters are 
allowed to reverse freely. 

These two criteria are conceptually simple and do not assume much about the evolution- 
ary process. Their differences can be appreciated by considering the transfomiation series 
"0, 1 , 2." If a particular tree topology "forces" the hypothesis that "0" evolves to "2," Wagner 
parsimony would add two steps to the tree, whereas Fitch parsimony would add only one 
step. Which result is biologically reasonable would depend on the justification for ordering 
or not ordering the transformation series. An important similarity of the two criteria is that 
both assign a cost to character reversals. Thus, there is no cost in temis of tree length if the 
root of the tree is changed. This is quite different from the characteristics of the next two 
parsimony criteria, where rerooting can have a considerable effect on tree length. 

3. Dollo parsimony (Farris, 1977) requires every synapomorphy to be uniquely 
derived, i.e., appearing only once on the tree. The synapomorphy may reverse, but once 
reversed it cannot reappear. Thus, parallel gains of apomorphies are prohibited. Dollo 
parsimony has been advocated for certain kinds of transfonnation series (e.g., endonuclease 
restriction site data; DeBry and Slade, 1985), but a "relaxed" Dollo criterion, which amounts 



56 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

to assigning a weight to the cost of reversal, might be more appropriate for these data 
(Swofford and Olsen, 1990). For example, you might assign a weight of "6" to gains of 
restriction sites, making their parallel evolution very costly in terms of tree length but not 
impossible. (The weight 6 is not arbitrary but refers to restriction sites six bases in length. 
The rationale is that the gain of a six-base site is six times less likely than its loss because a 
single nucleotide change would cause the loss.) 

4. Camin-Sokal parsimony makes the assumption that character evolution is 
irreversible. This is true, philosophically, because time is irreversible and the reversals are 
really new apormorphies. But we cannot know this a priori. This criterion is rarely used. 

Swofford and Olsen (1990) characterize general parsimony as an attempt to balance our 
knowledge of character evolution. You may have knowledge about different sorts of 
transformation series. This knowledge can be reflected in how you treat the characters. For 
example, you might have a data matrix composed of both morphological and biochemical 
characters. You may elect to treat a morphological transformation series of several characters 
as ordered because you have ontogenetic data that suggest a particular ordering. At the same 
time, you may have a number of alleles at a particular locus that you treat unordered because 
there is no clear evolutionary path connecting them that you can detemiine based on an 
independent criterion. This transformation series would be treated as unordered. You might 
have another transformation series comprised of the presence and absence of a particular 
restriction site and assign a weight to the presence. The result of such an analysis will be a 
tree supported by characters of various qualities. Tree topologies supported by characters 
given high weight are harder to reject than tree topologies whose branches are supported by 
characters weighted equally because of the cost of independently gaining high-weight 
characters. Thus, differential weighting should be carefully considered before application. 

Optimizing Trees 

You have already done some optimization in Chapter 2 when you produced trees that had 
alternate interpretations about the distribution of homoplasies (i.e.. Fig. 2. 1 1 ). Now, we will 
cover optimization in a more fonnal manner by considering two basic types, ACCTRAN 
and DELTRAN (Swofford and Maddison, 1987). ACCTRAN is equivalent to Farris 
optimization (Farris, 1970) where there is a known ancestor. The etymology of the name is 
derived from the fact that the procedure ACCelerates the evolutionary TRANsformation of 
a character, pushing it down the tree as far as possible. The effect is to favor reversals over 
parallelisms when the choice is equally parsimonious. The tree in Fig. 2.11c is an 
"ACCTRAN tree." hi contrast, DELTRAN DELays the TRANsformation of a character on 
a tree. The effect is to push the character up the tree as far as possible and to favor 
parallelisms over reversals when the choice is equally parsimonious. The tree in Fig. 2. 1 1 a is 
a "DELTRAN tree." 

ACCTRAN and DELTRAN operate under a general optimality criterion that is con- 
cemed with finding the most parsimonious character for each branch of the tree. Such 



TREE BUILDING AND OPTIMIZATION 57 

characters are parts of sets of characters termed most parsimonious reconstruction sets 
(MPR sets), which we will briefly discuss after considering ACCTRAN. The important 
thing to keep in mind is that both ACCTRAN and DELTRAN will yield the same results 
when there is no ambiguity, i.e., there are no equally parsimonious choices. Although 
DELTRAN may favor parallelisms, it is quite possible that if you select DELTRAN every 
homoplasy in your data set will be a reversal. In the following sections, we concentrate on 
transformation series that show reversals and/or parallelisms on particular tree topologies, 
but you should remember that characters showing no homoplasy are optimized in the same 
manner. 

ACCTRAN 

Farris (1970) pointed out that although a tree can be built with a Wagner algorithm, this 
does not always guarantee that the individual characters assigned to the hypothetical 
ancestors (represented by common nodes) will be optimal given the topology of the tree. 
Farris then provided an algorithm for optimizing these distributions. His algorithm includes 
transformation series with two to many characters. We will use a less set-theoretical 
description employing only binary transformation series. It will serve well enough to give 
you an idea of how this kind of optimization works. This type of optimization is called a 
"two pass" method (the Maddison et al. [1984] algorithm discussed in Chapter 3 is an 
example of a "one pass" method). Using the algorithm, you will assign certain characters to 
nodes in a pass from the terminal branches to the root (the downward pass) and then 
reevaluate these assignments in a pass from the root to the terminal branches (the upward 
pass). Transformation series of three or more states require a bit of knowledge about set 
theory. Those interested should consult Farris (1970) and Swofford and Madison (1987). 

We have broken the process down into three phases, the setup, downward pass, and 
upward pass. These rules work only in the binary case. 

1. The Setup. — On the tree, label each terminal taxon and each ancestor (Fig. 4.5a). 
Above the taxon name, place its character for a particular character transformation series 
(Fig. 4.5a). 

2. The Downward Pass. — Beginning with the terminal taxa and proceeding toward the 
root node, assign characters to the ancestral node according to the following rules. 

Rule 1, — If both terminal taxa have identical characters, label the node with that character 

(Fig. 4.5b, Rl). 
Rule 2. — If they have different characters, label the node with both characters (Fig. 4.5b, R2). 
Rule 3. — If one taxon (terminal or ancestral) has a single character (0) and another taxon 

(terminal or ancestral) has both characters (0,1), label their common node with the 

majority character (Fig. 4.5b, R3). If both characters are equally common (possible with 

polytomies), then assign both characters. 
Rule 4. — If both taxa have both characters (i.e., both are 0,1), then their common node will 

have both characters (see Fig. 4.6a). 



58 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



1 


1 








1 


1 1 





1 


A 

\ 


B 


c 

\ 


D 


E 

/ 


A B 


c 

\ 


D E 



a 



R3 




Fig. 4.5. — ACCTR AN optimization, a. The setup, b. The downward pass. c. The upward pass. R I -R6 are 
the application of the rules discussed in the text. 



3. The Root Node. — When you get to the root, label it with the ancestral character 
(whether it is or 1 or 0,1) regardless of what the last two taxa may indicate. In the "real 
world," this assignment would have been determined before you began by using the 
Madison et al. (1984) algorithm discussed in Chapter 3. We have arbitrarily selected "0" in 
these examples. 

4. The Upward Pass. — Begin working upward from the root node to the terminal taxa. 
The following rules apply to the upward pass. 

Rule 5. — If a character assignment at a node is 0, 1 . then assign the character of the node 
immediately below it (Fig. 4.5c, R5). (Equivalent to Rule 3, but going the other way.) 

Rule 6. — If a character assignment is either 1 or 0, then do not change that character 
assignment, even if the node below it is different (Fig. 4.5c, R6). 



TREE BUILDING AND OPTIMIZATION 



59 





A 



1 

B 





c 



1 

D 





E 





F 





Fig. 4.6. — State assignments on an ACCTRAN optimization pass. a. A downward pass. b. An upward pass. 



Example 4.3. — Simple character assignments. 

In Fig. 4.6a, we show the character assignments on the downward pass and in Fig. 4.6b 
the character assignments on the upward pass. Each assignment is associated with the mle applied. 

Example 4.4. — Complex character assignments (Fig. 4.7). 

As in Example 4.3, we illustrate character assignments for the downward pass (Fig. 4.7a) 
and the upward pass (Fig. 4.7b). 



60 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



110 110 10 1 




110 110 10 1 




Fig. 4.7. — Another example of ACCTRAN optimization, a. State assignments after the downward pass. 
b. Nodes optimized after the upward pass. 

ACCTRAN Exercises 

Use the information in the accompanying figures to optimize the character distributions. 
EXERCISE 4.7. — Use Fig. 4.8 for optimization and label the root node "0." 

110 




Fig. 4.8. — Tree for ACCTRAN optimization (Exercise 4.7). 



TREE BUILDING AND OPTIMIZATION 



61 



EXERCISE 4.8.— Use Fig. 4.9 for optimization and label the root node "0." 



110 10 1 




Fig. 4.9. — Tree for ACCTRAN optimization (Exercise 4.8). 



EXERCISE 4.9.— Use Fig. 4. 10 for optimization and label the root node "0. 



10 10 1 




Fig. 4.10. — ^Tree for ACCTRAN optimization (Exercise 4.9). 



Discussion 



ACCTRAN is a special case of a more general way to optimize characters on a tree 
(Swofford and Maddison, 1987). It will always insure that the length of the tree is minimized 
given a particular character matrix, but it will usually only show one interpretation of how 
the characters evolved (a formal proof of Farris's algorithm is provided by Swofford and 
Maddison [1987]). That is, it favors reversals over repeated origins when the choice is 
equally parsimonious. We can see this by examining the very simple tree in Fig. 4.11a. 

When you optimize the characters on the tree using ACCTRAN, the result shows a 
reversal from 1 to (Fig. 4. 1 1 b) (if this is not obvious, apply the ACCTRAN rules). The tree 
has a length of two steps. However, there is another equally parsimonious tree that interprets 
the evolution of a character coded "1" as a homoplasy (Fig. 4.11c). The problem is 
ACCTRAN will never allow you to find this tree. 



62 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



110 




110 






Fig. 4. 1 1 . — Four views of a tree. a. The tree with characters for the terminal taxa and the root node. b. The 
tree optimized by ACCTRAN. c. Another most parsimonious interpretation of the characters X and Y. d. The 
most parsimonious resolution (MPR) set. combining the information in b and c. 



Let US think about the character assignments of these two trees. Obviously, they differ. If 
we combine them into one tree, ancestor Y has a state set 0,1 rather than just 1 or (Fig. 
4.1 Id). Because this character set is the set that combines both possibilities, it is termed the 
"MPR set" for that particular ancestor (Swofford and Maddison, 1987). Because the MPR 
set contains two characters, we can assume that there are two equally parsimonious 
interpretations of character evolution. In Fig. 4.11b, the evolution of character 1 is 
accelerated: in Fig. 4.1 Ic, the evolution of 1 is delayed. 



Finding MPR Sets 

We find MPR sets by evaluating the possible character assignments for each interior node 
(i.e., each ancestor), which is accomplished by rerooting the tree at each of these nodes, 
beginning with the node closest to the root and working toward the terminal branches. 

1. Proceeding up the tree (Fig. 4.12a) from the root (X), reroot the tree at the first 
ancestral node (Y) (Fig. 4.12b). 

2. Optimize the character assignment of this node by using the downward pass as 
specified in ACCTRAN. The MPR set will be composed of the majority character using 
Rule 3. Note that neither character is in the majority, so both (i.e., 0, 1 ) are assigned to Y (Fig. 
4.12b). 

3. Proceed to the next ancestral node (Z) and reroot the tree (Fig. 4.12c). Follow step 2. 
To reduce redundant calculations, use the MPR set of any ancestor whose MPR set has 
already been detennined. In Fig. 4.12c, note that we have used the MPR set of Y as well as 
the character sets of C and D to detemiine the MPR set of Y. This is a short cut. You are 
finished after all ancestral nodes have been evaluated. 



TREE BUILDING AND OPTIMIZATION 



63 



A C 1 DO 



A 



X B 1 




A 



1 
B 



1 
C 




D 




X B1 Z 0,1 
Y 0,1 



Y 0,1 C 1 DO 





Z 0,1 



Fig. 4.12. — Finding the MPR set for ciade ABCD. a. Tree. b. Tree rerooted at Y and optimized usin^ 
ACCTRAN. c. Tree rerooted at Z and optimized using ACCTRAN. 



DELTRAN 

DELTRAN is the opposite of ACCTRAN. It favors parallelisms when given the chance 
to do so. To implement the DELTRAN option, you must be able to find the MPR sets for 
each ancestral node. 

1. Determine the MPR sets for each ancestral node by using the procedure outlined 
above, and place them on the tree. 

2. Implement DELTRAN as in the upward ACCTRAN pass, using Rules 5 and 6. 

Example 4.5. — MPR sets and DELTRAN optimization. 

We will use the tree in Fig. 4.13a to I) determine the MPR sets for each ancestor and 2) 
optimize the tree using DELTRAN. Note that the number of tenninal taxa used decreases as 
the MPR sets for their common nodes are included as we go up the tree. For example, A and 
OG are represented in Fig. 4. 1 3c by S and its MPR set (0). 

1 . We determine the MPR sets for each hypothetical ancestor, working from the root up 
the tree (Fig. 4.1 3b-g). 

2. We place the MPR sets on the original tree (Fig. 4.14a). 

3. Finally, we optimize the tree in an upward pass using Rules 5 and 6 (Fig. 4. 14b). 



DELTRAN Exercises 

EXERCISE 4.10.— Using the matrix in Table 2.2 (p. 19) and the tree in Fig 2. 1 1 a (p. 21 ), 
optimize transformation series 3 and 4 using DELTRAN optimization. Then optimize the 
tree using ACCTRAN. Compare your results with the trees in Fig. 2.11a, c. Use only the 
topology; disregard the characters on the tree. 



64 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 






1 








1 


1 








OG 


A 


B 


C 


D 


E 


F 


G 




CO D 1 

/ 

W 0,1 E 1 F GO 

/ \ 

V 1 xo 



B 



A 1 OG TO 



SO 



U 0,1 



CO D 1 



W 0,1 E 1 F 



V 1 



SO BO U 0,1 



T 



G 



XO 



CO D 1 



W 0,1 E 1 F GO 



TO VI X 



U 



CO D 1 



U El W 0,1 



V 0,1 



VI CO D 1 



UO F GO 



f w 0,1 9x0 

Fig. 4.13. — Finding the MPR sets (Example 4.5). a. Tree. b-g. MPR sets for each hypothetical ancestor. 



TREE BUILDING AND OPTIMIZATION 



65 






1 








1 


1 








OG 


A 


B 


c 


D 


E 


F 


G 







1 








1 


1 








OG 


A 


B 


C 


D 


E 


F 


G 




Fig. 4.14.— DELTRAN optimization using the MPR sets in Fig. 4.13 (Example 4.5). a. The original tree 
with MPR sets placed at the nodes, b. Each node optimized using DELTRAN. 



66 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

EXERCISE 4.11. — Use the tree in Fig. 4.15 to find the MPR set for each ancestor. Use 
these sets to optimize the tree with DEUTRAN. Then optimize using ACCTRAN. 








Fig. 4. 15. — Tree for MPR sets and DELTRAN optimization (Exercise 4.11). 



Current Technology 

Modem computer packages like PA UP (Swofford. 1990) are composed of programs 
containing a wealth of options. Actual analyses may take seconds, hours, or days. Interest- 
ingly, most of the time spent in analyzing the data is not spent on tree building per se but on 
evaluating tree topologies using a relevant optimality criterion. The approach to finding the 
optimal tree can differ considerably from traditional methods. Swofford and Olsen (1990) 
characterize three general approaches: 1 ) exhaustive searches, 2) branch-and-bound 
searches, and 3) heuristic searches. The first two can provide an exact soludon if the data 
matrix is small enough. By exact solution we mean that the resulting tree or set of trees will 
be the shortest tree(s) for the data given maximum parsimony as the criterion. The heuristic 
search will find a short tree(s), but there is no guarantee that the tree will be the shortest. 
Heuristic searches are performed on much larger data sets instead of using searches 
guaranteed to find an exact solution because for many large data sets there are no other 
alternatives (see below). The following abbreviated characterization of these three ap- 
proaches is abstracted from Swofford and Olsen ( 1990). 

1. Exhaustive search. — An exhaustive search consists of evaluating the data over all 
possible trees. Because there are no other possible topologies unevaluated, the shortest tree 
or set of trees will be found. As you will appreciate when you get to Chapter 6, the number 
of possible tree topologies increases at a very rapid rate. An exhaustive search on 10 taxa is 
very efficient, but the same procedure performed on 25 taxa may be prohibitive in terms of 
computer time. 

2. Branch-and-bound search. — Branch-and-bound algorithms can provide an exact 
solution for a larger number of taxa than exhaustive search can because the search procedure 
it employs has a provision for discarding trees without evaluating them if they meet certain 
criteria. Hendy and Penny (1982) first introduced branch-and-bound algorithms to phyloge- 
nedc analysis, and a good description of what they do can be found in Swofford and Olsen 



TREE BUILDING AND OPTIMIZATION 



67 



(1990). In essence, branch-and-bound works with a tree of possible trees termed a search 
tree (Fig. 4. 16). The root of the search tree is a three-taxon tree (unrooted so it is the only 
possible tree). From this root tree, other possible trees are derived by adding additional taxa 
at each place where they might be added. So, D could be added in one of three places to yield 
three trees, and E could be subsequently added in any one of five places to these three trees, 
yielding the 1 5 trees in Fig. 4. 16. Now, let us place an upper bound on the search. The upper 
bound is the maximum length a tree can obtain before it is eliminated. As we go from the 
root of the search tree towards its tips, we will only reach the tips along a certain path if our 
upper bound is not exceeded. If we exceed the upper bound, we cut off that branch and do 
not evaluate any additional branches connected to it. Rather, we backtrack down the search 
tree and proceed up another branch. If we find a tree that is better than the upper bound (i.e., 
a tree shorter than the one we used for the initial upper bound), we adopt this superior 



B E 



B E 




Fig. 4. 16. — Trees illustrating the branch-and-bound algorithm. Redrawn from Swofford and Olsen (1990). 



68 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

Standard for all subsequent analyses. Thus, we might eliminate additional trees that would 
be acceptable given the old, more relaxed, upper bound. Trees that are not eliminated are 
candidates for status as most parsimonious trees. If a better upper bound is found, these trees 
are subject to further evaluation. What is left is the set of optimal trees. This is the trick: by 
not evaluating all the branches to all the tips, we avoid an exhaustive search without 
suffering loss of precision because we do not have to evaluate tens, hundreds, thousands, or 
even millions of possible tree topologies that are not close to the optimal solution. Obvi- 
ously, we need to pick an initial upper bound before the analysis begins. This can be done by 
picking a random tree and calculating its tree length. For our example of five taxa (Fig. 4. 16), 
we would pick any of the 15 trees, calculate its tree length, and begin. If that tree turned out 
to be the worst possible tree, then we would expect to quickly find a new, and lower, upper 
bound. If the tree turned out to be the most parsimonious tree (a 15-to- 1 shot), then the upper 
bound would be so low as to quickly eliminate many of the less parsimonious trees. PAUP 
uses the next category of methods, heuristic methods, to pick a near opfimal tree whose 
length is used as the initial upper bound. Tliese and other tricks speed up the process 
(Swofford, 1990; Swofford and Olsen, 1990). 

3. Heuristic search. — Heuristic search begins by building a tree. One common way to 
do this is via the Wagner algorithm. A number of other approaches can also be used, 
depending on the computer package. The tree obtained may be optimal, especially if the data 
are relatively free of homoplasy. However, it may not be optimal. The manner in which taxa 
are added to the growing tree constrains the topologies that can be built from taxa added 
subsequently, ending in a "local optimum," i.e., a tree that is the "optimum" given the 
constraints provided by the way the taxa were added but not necessarily the best tree if all 
possible trees were evaluated. This situation is rather analogous to proceeding along a 
reasonable path in a search tree but not having the option of backtracking to try other 
possibilities. Heuristic search routines attempt to circumvent local optima by branch 
swapping. Branch swapping involves moving branches to new parts of the tree, producing 
new tree topologies. The data are optimized on the new topology, and if the tree is shorter, it 
is subjected to additional branch swapping. Eventually, these rounds of branch swapping 
will lead to an optimal tree or set of optimal trees. Several pitfalls can be encountered, and 
strategies for avoiding them must be adopted (Swofford and Olsen, 1990). 

Chapter Notes and References 

1. Hennig argumentation is covered in some detail by Hennig (1966), Eldredge and 
Cracraft (1980), and Wiley (1981a). It works well with relatively small data sets with little 
homoplasy. However, we continue to be amazed to see how few homoplasies it takes in a 
data matrix to yield more than one most parsimonious tree. 

2. Those akeady familiar with data analysis will note that we have not discussed 
alternative approaches to phylogeny reconstruction such as compatibility analysis, analysis 
of distance data (Farris, 1972, 1981, 1985, 1986; Felsenstein, 1984, 1986; HiUis, 1984, 



TREE BUILDING AND OPTIMIZATION 69 

1985), likelihood methods (Felsenstein, 1973a,b, 1981; Thompson, 1986), and bootstrap 
methods (Felsenstein, 1985; Sanderson, 1989). You should consult the literature about these 
controversies. 

3. Those interested in analysis of molecular data should consult Hillis and Moritz (1990). 

4. Finding the MPR set for transformation series with more than two characters is not so 
easy and requires more set manipulations. Swofford and Maddison (1987) outline the 
procedure and provide proofs of both Farris's (1970) optimization algorithm and their own 
algorithms. However, the paper is very technical. Farris (1970) gives a clear example of how 
to optimize a multicharacter transformation series using ACCTRAN. 

5. Remember that ACCTRAN and DELTRAN will give the same solutions for character 
evolution only if it is equally parsimonious to do so. The majority of transformation series in 
a particular data matrix may not contain MPR sets with two or more characters, and both 
ACCTRAN and DELTRAN will give the same values. Also, consider the situation where 
we may have many equally parsimonious decisions to make, each involving a different 
transformation series. We might prefer one transformation series to be interpreted as being 
subject to parallel evolution (DELTRAN) while another is considered as being subject to 
reversals (ACCTRAN). In any one analysis, however, we cannot "mix and match." Given a 
particular topology, we can only opt for a uniform DELTRAN or a uniform ACCTRAN 
optimization. The extent to which a particular investigator opts for the interpretation that a 
particular transformation series is "more likely" to be subject to parallelism than to reversal 
does not depend on an optimization routine but on specific assumptions about the evolution- 
ary constraints placed on the descent of particular kinds of characters. 



70 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 1 9 



Chapter 5 

TREE COMPARISONS 



In the early analyses of groups using phylogenetic techniques, only rarely did investiga- 
tors report more than one tree, frequently because the data were "clean," i.e.. relatively free 
from homoplasy. but sometimes because the investigator did not take the time to explore the 
data set for additional trees. Computers have changed this. A computer will take the time to 
find as many alternative trees as the program it employs can identify. As a result, more trees 
are being reported, and measures are needed to sort them out. In this chapter, we will 
introduce you to 1) some basic tree measures and 2) consensus techniques, i.e.. techniques 
for exploring the similarities and differences between trees with differing branching stmctures. 

Summary Tree Measures 

Summary tree measures are designed to give you a set of basic "statistics" you can use to 
evaluate the differences between two trees that have been generated from the same data set. 
They are relative measures that say something about the basic structure of the data, the 
optimization technique employed, and the algorithm used. They are not very useful for 
comparing the results of two different data sets for the same taxa. 

Tree Length 

We have been working with tree length since Chapter 2. Tree length is calculated by 
summing the number of character changes along each branch and intemode of the tree. For 
a given data set, the "best tree" is defined as the tree of shortest length because, as we 
discussed in Chapter 2, it provides the most parsimonious description of all homology and 
polarity aiguments taken together. We hope that the shortest tree will also be the best 
estimate of the actual common ancestry relationships of the taxa analyzed. Two or more 
shortest trees represent equally parsimonious solutions to the same data set for a particular 
analysis. There are two kinds of equally parsimonious trees: 1 ) the set of trees that show the 
same common ancestry relationships but differ in character inteipretation and 2) the set of 
trees that differ in topology and thus represent different views of the common ancestry 
relationships. Tliese two types of equally parsimonious trees have different qualities. For 
systematics and biogeography, equally parsimonious trees that differ in character interpreta- 
tions but have identical topologies do not affect any subsequent analysis that takes advan- 
tage of the tree topology. However, the differences among these trees might be important 
when dealing with tests of evolutionary mechanisms and character evolution. Trees of 
differing topology directly affect subsequent comparative analysis and can lead to problems 
in presenting taxonomies because there are at least two different views of the common 
ancestry relationships among the taxa involved in the analysis. 



72 



KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 



Consistency Indices 

Kluge and Farris ( 1969) introduced consistency indices as measures of how transforma- 
tion series and entire data matrices "fit" particular tree topologies. Nontechnically, transfor- 
mation series with little or no homoplasy have high consistency index values ( 1 .0 is the 
highest value possible), whereas those that show considerable homoplasy have low values. 
Aside from tree length, consistency indices are the most commonly reported values for trees. 
Since their introduction in 1969, several modified consistency indices have been suggested. 
We will cover the basics of these measures using the data matrix in Table 5.1 and the trees in 
Fig. 5.1. We will first cover the measures for individual transformation series and then 
consider the ensemble measures for entire data matrices. 



Table 5. 1 . — Data matrix for the hypothetical clade A-E and its sister group OG. 











Transformation 


series 








Taxon 


1 


2 


3 


4 


5 


6 


7 


8 


OG 


























A 
















1 





1 


B 




1 


1 





1 





1 





C 







1 


1 


1 











D 




1 


1 





1 


1 


1 





E 




1 


1 


1 


1 


1 


1 


1 



The consistency index of a transformation series of discrete characters (c or ci) is the ratio 
of the minimum amount of changes (steps) it might show (m) and the amount of change 
(steps) it does show on a particular tree (s): 



c = 



m 

s 



For binary transfomiation series, m = 1 ; for multicharacter series, the m-value will equal 
the minimum number of steps possible. For example, a transfonnation series with three 
characters (0, 1, and 2) would evolve a minimum of two steps. For simplicity, we use only 
binary transformation series in both the examples and the exercises. 

Consider TS 1 of Table 5. 1 . As a binary transfomiation series, m = 1 . Now look at Fig. 5.1. 
There has been one transfonnation from 1-0 to 1-1 and no reversals from 1-1 to 1-0. The 
number of steps (or length) of this series on the tree is s = 1. The consistency index for this 
transfonnation series is 



c = 



m 

s 



1.0. 



Now consider TS 8. As a binary transfonnation series, m = 1 . On the tree, however, 
character 8-1 has arisen twice, so s = 2 and 



c = m = 1 . 0.50. 

s 2 



TREE COMPARISONS 



73 



OG 



D 



a 





OG A B C D E 





Fig. 5.1. — a. Tree for the hypothetical clade A-Eand its sister group OG. b. c. Polytomies forTS 2. 
d. Poly tomy for TS 3. 



Now consider TS 6. another binary character with s-\. Figure 5.1 shows that 6-0 has 
given rise to 6-1 twice, once in taxon A and once in the ancestor of taxa D and E. The 
consistency ijidex for this character is 



c = m = 1 = 0.50. 

s 2 



Although the consistency indices for TS 6 and TS 8 are identical, there is a difference in 
their performance on the tree. TS 8 shows total homoplasy in the evolution of character 8-1. 
whereas TS 6 shows only partial homoplasy in the evolution of 6-1. That is, a derived 



74 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 1 9 

character in TS 6 does lend support to the tree topology in Fig. 5.1, whereas derived 
characters in TS 8 lend no support to the tree topology at all. Yet, they have exactly the same 
consistency index values. 

To overcome this problem. Farris (1989b) introduced the rescaled consistency index 
(actually first introduced in Farris' Hennig 86 [1989a]). Tlie rescaled consistency index (re) 
is the product of the original consistency index (c) of Kluge and Farris (1969) and the 
retention index (r) of Farris (1989b). The retention index measures the fraction of apparent 
synapomorphy to actual synapomorphy. To calculate the retention index we need the s- 
value, the m-value, and a new value called the g-value, for each transformation series. The g- 
value is a measure of how well a transfonnation series might perform on any tree, i.e., how 
many steps would it take to explain evolution within the transformation series under the 
worst possible condition. The case of any tree is the case of the phylogenetic "bush," a 
polytomy involving all taxa (including the outgroup). 

Consider TS 2 in Table 5. 1 . In the worst possible circumstances (the phylogenetic bush), 
what is the minimum number of steps needed to explain the evolution of the characters in 
this transformation series? In Fig. 5. lb we have a bush (including the outgroup). If we assign 
2- 1 to the root, then 2-0 would evolve three times, creating three steps. If we assign 2-0 to the 
root, then 2-1 requires three steps (Fig. 5.1c). So, assigning 2-0 or 2-1 to the root results in 
the same number of steps, and g = 3. Any fewer number of steps would require us to group 
taxa, and this would not represent the minimum number of steps in the worst case (the 
g-value) but would instead represent the minimum number of steps in the best case (the 
m-value). 

Now consider TS 3. If we assign 3-1 to the root (Fig. 5.1a), then 3-0 has to evolve twice 
(g = 2). If we assign 3-0 to the root, then 3- 1 has to evolve four times. Parsimony prefers two 
steps to four steps, so the best we can do under the worst tree topology is two steps. 

For binary characters we can apply a short-cut to detennine the g-value. For a transforma- 
tion series with binary characters, the g-value is the smallest of the values for the occurrence 
of the two characters. TS 3 contains two 3-0 values and four 3-1 values, and g = 2. The 
number of 0-values and 1 -values in TS 2 is equal, so the g-value equals this value (g = 3). 
There is only a single 1 -0 in TS 1 , so g = 1 . 

The retention index, r, is defined by Farris (1989b) as 

r = g - --=^. 3-1:^ [ice Bn^trXA-l 

When we calculate some of the r-values for the transformation series in Table 5.1 using 
the tree in Fig. 5.1a, we find the following values: 

TS 2: r = 
TS 6: r = 
TS 8: r = 



3-1 . 
3-1 


tolto 

II 


3-2 . 
3-1 


= ^ = 0.50 


2-2 
2-1 


= ^ = 0.00. 



TREE COMPARISONS 75 

Note the contrast between TS 6 and TS 8. Although there is no actual synapomoqDhy in 
TS 8, there is apparent synapomorphy (i.e., reflected by the original coding of the homopla- 
sies represented by the code "8-1"), and the retention index for this transformation series is 
zero. Some synapomorphy is present in TS 6, so its retention index is greater than zero. 

The rescaled consistency index (re) for a transformation series is the product of the 
consistency index and the retention index: 

TS2: rc = (l)(l)= l.(X) 
TS 6: re = (0.5)(0.5) = 0.25 
TS8: re = (0)(0.5) = 0.0. 

Note the relative performance of the two transformation series showing homoplasy. 

Finally, consider TS 1 . This transformation series presents the special case m = s = g = 1 . 
Obviously chai-acter 1 - 1 represents a synapomorphy for the group given the assumption that 
1-0 is plesiomorphic. But what if 1-1 were plesiomorphic and 1-0 were autapomorphic for 
the outgroup? There is no way to specify this, so we will calculate r = and re = 0. 
Biologically, we "know" that 1-1 is a synapomorphy, and thus r = 1 and re = 1.0. But 
algorithmically, the only way to produce this result would be to include another outgroup. 
Biologically, we can interpret this transformation series directly. A character shared by all 
members of the ingroup is consistent with all possible topologies and thus is not informative. 
Likewise, autapomorphies will yield r = and re = because they are also consistent with all 
possible topologies (see Exercise 5.1). 

Ensemble Consistency Indices 

Ensemble consistency indices can be used to examine the relationship between an entire 
data matrix and a particular tree topology. The most commonly used index is the ensemble 
or "overall" consistency index (CI) of Kluge and Farris (1969). which can be calculated very 
simply for a binary matrix by taking the ratio of the number of data columns and the length 
of the tree. In general, a high CI indicates that the data matrix "fits" the tree well (i.e., 
contains little homoplasy for the particular tree topology), whereas a low CI does not. 

Although the CI is the most commonly reported measure of fit between a character matrix 
and a tree (Sanderson and Donoghue, 1989), it suffers from some problems. Brooks et al. 
( 1986) showed that the CI was influenced by the number of autapomorphies that are present 
in the data matrix. The actual support for a particular tree may be less than the apparent 
support because autapomorphies (c = 1.0) contribute to the CI without supporting any 
particular tree topology. One solution to this problem would be to eliminate autapomorphies 
from the calculations (Carpenter, 1988). Another problem is the negative relationship 
between the CI and the size of the data set (Archie, 1989), rendering the CI suspect when 
comparing different groups of taxa or the same taxa for different data matrices of various 
sizes. 

In an effort to address some of these problems, Farris (1989b) suggested that ensemble 
consistency indices be calculated using rescaled values. These calculations are demon- 
strated in Example 5.1. 



76 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



Example 5.1. — Rescaled consistency indices. 

We will use Table 5. 1 and Fig. 5. 1 a as the basis for calculating the ensemble values of the 
retention index (R) and the rescaled consistency index (RC). To do this, we obtain sums of 
the S-, m-. and g-values for each individual transformation series. These values are shown in 
Table 5.2. Note that because we are working with binary transformation series, m = 1 for 
each data column and thus M = 8 (the number of data columns). If we sum each s-value, we 
arrive at S = 1 1 , the length of the tree. The sum of the minimum number of steps for each 
transformation series given a polytomy (the g-value) yields G = 18. We can now calculate 
the CI, R, and RC: 

M 8 



CI 



11 



= 0.727 



R = 



G-S 18-11 



= 0.700 



G-M 18-8 
RC = (0.727) (0.700) = 0.509. 



Table 5.2.- 


—Some 


values* 


used to calculate rescaled 


consistency indices (Example 5 


1). 


TS 


m 




s 


g 


ci 


r 




re 


1 






1 


1 


1.00 


0/0 




0/0 


2 






1 


3 


1.00 


1.00 




1.00 


3 






1 


2 


1.00 


1.00 




1.00 


4 






2 


2 


0.50 


0.00 




0.00 


5 






1 


2 


1.00 


1.00 




1.00 


6 






2 


3 


0.50 


0.50 




0.25 


7 






1 


3 


1.00 


1.00 




1.00 


8 






2 


2 


0.50 


0.00 




0.00 


Totals 


8 




11 


18 











* m = no. changes a character might show on a tree; s = no. changes a character does show on a tree; g = minimum 
no. steps for each TS given a polytomy; ci = character consistency index: r = character retention index; re = character 
rescaled consistency index. 

The RC excludes characters that do not contribute to the "fit" of the tree, preventing 
inflation of the CI. The RC is superior to the method of excluding autapomorphies because 
it not only excludes these characters, but also excludes totally homoplastic characters, 
preventing them from artificially inflating the measure of fit (e.g., TS 8), while allowing 
characters that are partly homoplastic but partly supportive of the tree topology (e.g., TS 6) 
to contribute to the ensemble value. 



The F-Ratio 

The F-ratio (apparently first presented in the LFIT function of PHYSYS, a computer 
program developed in 1982 by M. F. Mickevich and J. S. Farris) is another commonly 



TREE COMPARISONS 77 

reported tree statistic. It is a measure of the differences between the phenetic differences 
between taxa and the path-length distances between these taxa as shown by a particular tree. 
Low F-ratios are considered superior to high F-ratios. To calculate the F-ratio we need 1 ) a 
data matrix. 2) a matrix of phenetic differences, and 3) a matrix of path length (patristic) 
distances. The F-ratio is then calculated in the following manner. 



't 



Example 5.2. — F-ratio calculations. 

1. From the data matrix (Fig. 5.2a), solve the phylogenetic relationships of the group 
(Fig. 5.2b). 

2. From the data matrix, prepare a matrix of phenetic distances (Fig. 5.2c). We do this by 
constructing a taxon x taxon matrix. The phenetic difference between two taxa is the sum of 
their absolute differences. So, the distance from OG to A is the difference between the OG 
character vector and the A character vector, which is 2. Tlie distance between B and D is 3, 
etc. 

3. From the phylogenetic tree, prepare a matrix of path-length (patristic) distances (Fig. 
5. 2d) by preparing another taxon x taxon matrix. This time we count the steps along the path 
specified by the tree. For example, there are two steps from OG to A and from A to B. The 
difference between the phenetic and path-length distances becomes apparent when we 
compare the A to D path. Phenetically. the difference between A and D is 3. Patristically, the 
path-length difference is 5 because the tree interprets character 5-1 as a homoplasy. 

4. Calculate the /-statistic. The /-statistic is the sum of the difference matrix. The 
difference matrix is a matrix of differences between the phenetic and path-length matrices 
(Fig. 5.2e). For example, there is no difference in the A/B cell, so the value is "0," whereas 
there is a difference of 2 in the A/D cell, so the value is "2." 

5. The F-ratio is calculated as the ratio of the /-statistic and the sum of the phenetic 
distance matrix. To sum the phenetic distance matrix, you simply sum each column and then 
add the column sums together. In this case, the F-ratio is 



F-ratio = ( f-''^'''''' )(100) 

patristic matrix 

:. (— )(100) 

^ 28 ^ 

= 7.14. 



The F-ratio has two major drawbacks. First, it is not entirely clear what the F-ratio is 
supposed to measure because we do not really know what a phenetic difference means 
relative to a phylogenetic tree. Second, although the F-ratio can distinguish autapomorphies 
from intemal synapomorphies (i.e., synapomorphies not associated with the root intemode), 
it treats synapomorphies at the root intemode as if they were autapomorphies because the 
calculation of the F-ratio is independent of the placement of the root. 



78 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



a 



TS 



Taxon 


1 


2 


3 


4 


5 


OG 

















A 


1 











1 


B 


1 


1 











C 


1 


1 


1 


1 





D 


1 


1 


1 


1 


I 



step 2 




Step 3 



Taxa 



Taxon OG A B C D 



OG 











A 


T 









B 


-> 


2 







C 


4 


4 


T 





D 


5 


3 


3 


1 



Taxa 



Taxon OG A B C D 



OG 









A 


2 







B 


2 


2 





C 


4 


4 


") 


D 


5 


5 


3 



Step 4 



Taxa 



Taxon OG A B C D 







OG 











A 












B 













C 














D 





2 









Fig. 5.2. — Matrices and tree illustrating F-ratio calculations (Example 3.2). a. Data matrix, b. Tree. 
c. Phenetic distance matrix, d. Patristic distance matrix, e. Difference matrix. 



Tree Summaries Exercises 



EXERCISE 5.1.— Using the data matrix in Table 5.3, derive the tree and calculate the tree 
length, CI, R. and RC. 



TREE COMPARISONS 



79 



Table 5.3. — Data matrix for calculating summary statistics (Exercise 5.1 ). 



Transformation series 


Taxon 


1 


2 


3 


4 


5 


6 


7 


8 


X (outgroup) 


























A 


1 


1 

















1 


B 


1 


1 


1 


1 














C 


1 


1 


1 


1 


1 


1 


1 






EXERCISE 5.2.— Using the matrix in Fig. 5.2a, calculate the tree length, CI, R, and RC for 
the tree in Fig. 5.2b. 

EXERCISE 5.3.— Calculate the tree length, CI, R, and RC for the taxa and characters 
shown in Table 5.4 and Fig. 5.3. 

Table 5.4. — Data matrix for calculation of tree length, CI, R. and RC from Fig. 5.3 (Exercise 5.3). 











Transformation series 








Taxon 


1 


2 


3 


4 


5 


6 


7 


8 


OG 


























A 


1 











1 


1 


1 


1 


B 





1 








1 


1 


1 





C 








1 








1 


1 





D 











1 








1 


1 




Fig. 5.3. — The tree used to calculate tree length, CI, R, and RC (Exercise 5.3). 



80 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



EXERCISE 5.4. — Derive the tree and calculate the tree length, CI, R, and RC for the taxa 
and characters shown in Table 5.5. 



Table 5.5. — Data matrix for calculation of tree length. CI. R. and RC (Exercise 5.4). 



Transformation series 


Taxon 


1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


Outgroup 
































M 


1 


1 








1 


1 


1 


1 








N 


1 


1 


1 


1 


1 


1 


1 





1 





O 


1 


1 


1 


1 

















1 



EXERCISE 5.5. — Derive the tree and calculate the tree length, CI, R, and RC for the taxa 
and characters shown in Table 5.6. 



Table 5.6. — Data matrix for calculation of tree length, CI, R, and RC (Exercise 5.5). 













Transformation 


series 










Taxon 


1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


11 


OG 



































A 
































I 


B 




1 










1 


1 














C 




1 










I 


1 








1 


1 


D 




1 




1 


1 


1 





1 


1 








E 









1 


1 








1 


1 








F 




1 
















I 


I 





I 



Consensus Techniqiies 

One easy way of dealing with a set of topologically different but equally parsimonious 
trees or a set of topologically different trees derived from different data sets is to combine 
their information in some manner. Consensus trees are trees that combine the infoimation, 
or knowledge claims, about grouping contained in two different trees into a single tree. As 
such, consensus trees must be used very carefully. It would be tempting, for example, to 
decide that a consensus tree, containing all the information from two different but equally 
parsimonious trees, gives you the best estimate of the phylogenetic relationships among 
groups. However, this is not true. In this section, we will cover some of the basic consensus 
techniques and give you some guidelines for using them. Like the Wagner algorithm section 



TREE COMPARISONS 



81 



of Chapter 4. we do not expect that you will use the techniques presented here to generate 
consensus trees because the computational effort for even simple trees is rather great, 
histead, these exercises are designed to give you a concept of what these trees do to groups 
elucidated by parsimony. We will cover three kinds of trees that can be generated by 
different consensus methods. Each can be used in different situations to answer different 
questions. 

Strict Consensus Trees 

Strict consensus trees (Sokal and Rohlf, 1981) contain only those monophyletic groups 
that are common to all competing trees. The ellipses in a Venn diagram can represent these 
groups. (We shall make extensive use of Venn diagrams in this and following sections, so 
you might want to look back at Chapter 1 for a quick review, then try the Venn diagram 
Quick Quiz.) 

Example 5.4. — A strict consensus tree. 

The two trees shown in Fig. 5.4 have different knowledge claims about the relationships 
of taxa C, D, and E. These trees are logically inconsistent, but they do contain some common 
knowledge claims. 



B 



B 



D 



a 





Fig. 5.4. — Two hypotheses of the relationships among taxa A-E (Example 5.4). 



We construct a strict consensus tree for these alternative trees in the following manner. 

1 . Draw Venn diagrams for each tree (Fig. 5.5a, b). 

2. Combine the Venn diagrams into a single diagram (Fig. 5.5c). It should come as no 
surprise that the groupings intersect. 

3. Erase all intersecting ellipses (Fig. 5.5d). The result is a consensus Venn diagram. 

4. Translate the consensus Venn diagram into a stiict consensus tree (Fig. 5.6). 



82 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 





c 



B 



(^^ 



c^o 




B 



3 



D 



Fig. 5.5. — Obtaining a strict consensus Venn diagram, a, b. Venn diagrams of the original trees in Fig. 5.4. 
c. Combined Venn diagrams, d. Strict consensus Venn diagram. 



EXERCISE 5.6. — Using the trees shown in Fig. 5.7, constnict a strict consensus tree using 
Venn diagrains. 



Example 5.5. — Another strict consensus tree. 

Our first example concerned trees that contained logical inconsistencies. Strict consensus 
trees, however, also deal with different trees, or parts of trees that do not have logically 
inconsistent topologies. In such cases, the strict consensus tree would be the tree of lowest 
resolution. Consider the trees in Fig. 5.8a-c. If we employed Venn diagrams, we would see 
that these trees are logically consistent with each other. Tlie strict consensus tree (Fig. 5.8d) 



TREE COMPARISONS 



83 



B 



D 




Fie. 5.6. — A strict consensus tree of the taxa A-E derived from Fis. 5.5d. 



B 



B 



D 





Fig. 5.7. — Phylogenetic trees for construction of strict consensus tree (Exercise 5.6). 

represents the "lowest common denominator" concerning the knowledge claims of the three 
trees. For example, in moving A to the root node, we cannot maintain B and C as a 
monophyletic group, even though two of the three trees (5.8a, b) contain this grouping. 
Why? Because the third tree (5.8c) makes the claim that A, B, and C fomi a monophyletic 
group. 



Adams Trees 

Adams consensus trees are designed to give the highest "resolution" possible between 
two or more trees. When these trees are logically inconsistent, the taxa responsible for the 
conflict are relocated. Thus, Adams consensus trees do not necessarily reflect monophyletic 
groups that are supported by the original data matrix. A detailed desciiption of Adams trees 
can be found in Adams (1972). He presents two cases, trees with labeled nodes (e.g., 
specified ancestors) and trees with unlabeled nodes (trees of common ancestry). We will 
deal only with the second case, trees with unlabeled nodes. 

Adams techniques are built around several steps. The investigator partitions the taxa into 
sets across each of the trees, beginning with the root node. Tliis partitioning groups taxa into 
sets based on their connection via a branch to the root node. For example, the partition sets 



84 



KU MUSEUM OF NATUR.'U. HISTORY, SPECIAL PUBLICATION No. 19 



B C A 



B C A 





B C A 



B 



X 





Fig. 5.8. — Three trees (a-c) and their strict consensus tree (d). 

for Fig. 5.8b are { X j and { ABC } . the sets of taxa connected to the root via a branch. (Note 
that { ABC ) is not subdivided into { A } and { BC } at this time. Such a subdivision would 
happen only in the next round when the node common to these taxa alone is considered.) 
The investigator then compares these sets across all trees to find partition products, which 
are the sets (=taxa and/or monophyletic groups of taxa) common to each of the original trees. 
For example, the tree in Fig. 3.8a has the partition sets { BC } , { A } , and { X } . Tlie partition 
products of the trees 5.8a and 5.8b are [X], [A], and [BC]. The set { ABC } is not a product 
because it does not appear in the partition sets common to both trees. Note that single taxa as 
well as sets of taxa can be products. However, single taxa cannot be further partitioned, and 
sets with only two taxa are automatically partitioned into only the two single taxa. Note that 
nonempty partition products require at least one taxon to be part of the partition sets 
compared between trees. Finally, the method works from the root of the original trees 
outwards, identifying partition sets until the temiinal taxa are reached. Any taxa partitioned 
out are not considered at higher nodes even if they are parts of monophyletic groups in one 
or more of the original trees. 



Example 5.6. — For this example, we will use the phylogenetic trees shown in Fig. 5.8a-c. 
Because the trees contain only four taxa, we examine only two nodes; the second node we 
examine contains only two taxa. 

1. Partition the taxa into partition sets connected to the root node. This yields the 
following partition sets for each tree. 



TREE COMPARISONS 



85 



Tree 5.8a (3 partition sets): { X 1 , { A 1 . j BC ) 
Tree 5.8b (2 partition sets): { X ) , { ABC 1 
Tree 5.8c (2 partition sets): { X } , { ABC } 

2. Determine the partition products by finding the intersections (common set elements) 
of each partition set that yields a nonempty partition product. The simplest of these partition 
products is the intersection of the sets { X ) . j X } . and { X } , yielding [X] as the partition 
product. Another is formed from the intersections of { A } , { ABC } . and { ABC ) . yielding [A] 
as the partition product. The final product is formed by the intersections of j BC | . j ABC }, 
and {ABC), yielding [BC] as the partition product. 

3. From the root, proceed to the next node that contains more than one taxon and repeat 
the partitioning and determine the partition products as in steps 1 and 2. Repeat up the 
branch or along sister branches until all terminal taxa are partitioned. In this example, there 
is only one final node that needs to be considered, the node leading to { BC | in tree 5.8a and 
to {ABC) in trees 5.8b and 5.8c. Note that (A) has already been paititioned, so the only 
products of partition are [BC] for each of the three trees. 

4. Form the Adams consensus tree from the sets detennined by partitioning, beginning at 
the root and using the partition products as sets. Figure 5.9a shows the Adams consensus tree 
derived from step 4. Tlie first partition products, [A], [X], and [BC], are joined at the root 
node. The set { BC ) is automatically paititioned (there are only two terminal taxa), yielding 
the dichotomy (Fig. 5.9b). 



BC 



B 



X 





Fig. 5.9. — Adams consensus tree for Example 5.6. a. Result of adding the first partition products, b. Final 
Adams consensus tree. 



Example 5.7. — For this example, we will use the trees shown in Fig. 5.10a-c. Tliese trees 
contain more taxa and nodes and will allow us to form some additional partition products 
that are not trivial (i.e.. that contain more than one or two taxa). 

1 . Form the first partition sets for the three trees. Note that the partition sets of trees 5 . 1 Oa 
and 5.10c are exactly the same, although the hypotheses of common ancestry within these 
groups are different. 



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KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



M 



N O P Q 



M 



Q 



NOP 





M 



N O P Q 




Fig. 5.10. — Trees for construction of Adams consensus tree (Example 5.7). 



Tree 5.10a (2 partitions): 
Tree 5.10b (2 partitions): 
Tree 5.10c (2 partitions): 



{M}.{N0PQ1 

{MQ},{NOPl 
{M}, {NOPQl 



2. Determine the partition products. 



Partition product 1 
Partition product 2 
Partition product 3 



|M},{MQ}.{M} = [M] 

I NOPQl. I NOP I. {NOPQ) = [NOP] 

|NOPQl.{MQ}.{NOPQ} = [Q] 



Note that the partition set {NOPQ} of both trees 5.10a and 5.10c have been used twice 
because one element (Q) appears in a different section of Fig. 5. 10b and the other elements 
appear in a different set { NOP 1 . Also note that other combinations are possible, but they 
contain no interesting taxa. For example, the intersection of {M) from tree 5.10a and 
(NOPQ) from 5.10c yield an empty set. 

3. Proceed to the next node up from the root of each tree and determine the partition sets, 
deleting any taxa that might have already been partitioned as single taxa. 

Tree 5.10a (2 partitions): {N}, {OP} 
Tree 5.10b (2 partitions): {Nj, {OP} 
Tree 5.10c (3 partitions): (N), {O), {P) 

Note that Q does not appear in either tree 5.10a or 5.10c because it has already been 
partitioned. As we will see. Q will join the Adams tree at the root because it is a partition 
product of the root node. 

4. Fonn the partition products. 



TREE COMPARISONS 



87 



Partition product 1 
Partition product 2 
Partition product 3 



|N1.{N1.|N1 = [N] 
{OP}.{OP).{0} = [0] 
{OPl.{OPl,{P| = [P] 



5. All terminal taxa have been partitioned. Form the consensus tree, beginning from the 
root and connecting each product of the partition associated with each node (Fig. 5.11). 

M Q N O P 




Fig. 5.11 . — Final Adams consensus tree for Example 5.7. 



EXERCISE 5.7. — Using the trees in Fig. 5.12, construct an Adams consensus tree. 



B 



D 



B 





Fig. 5.12. — Trees for construction of Adams consensus tree (Exercise 5.7). 



EXERCISE 5.8. — Using the trees in Fig. 5.13, construct an Adams consensus tree. 



B 



D 



D 




a 




Fig. 5.13. — Trees for construction of Adams consensus tree (Exercise 5.8). 



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KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



Majority Consensus Trees 

Majority consensus trees operate on a "majority rule" basis (Margush and McMorris, 
1981). therefore the consensus tree may be logically inconsistent with one or more of the 
original parsimony trees. If you have one tree that hypothesizes that A and B are more 
closely related to each other than either is to C (Fig. 5.14a) and two trees that hypothesize 
that B and C are more closely related to each other than either is to A (Fig. 5. 14b, c), then the 
majority consensus tree will have the latter topology (Fig. 5. 14d). If there is an "even vote," 
then the result is a polytomy. If the trees are logically consistent, then the most resolved tree 
is preferred (just as in the case of Adams consensus trees). 



B 




B 




B 




abed 

Fig. 5. 14. — Three trees (a-c) and the majority consensus tree (d) for taxa A-C. 



B 




EXERCISE 5.9. — Use the principles presented above to construct a majority consensus 
tree for the trees shown in Fig. 5.15. 



Y X W V 



Y X W V 





X 



W 



V 




Fig. 5.15. — Trees for construction of majority consensus tree (Exercise 5.9). 



TREE COMPARISONS 



89 



EXERCISE 5.10. — Using the trees in Fig. 5.16, construct a strict consensus tree, an Adams 
consensus tree, and a majority consensus tree. 



ABCDEFGH 



ABCDHEFG 





ABCDEFGH 




Fig. 5.16. — Trees for construction of strict, Adams, and majority consensus trees (Exercise 5.10). 



Chapter Notes and References 

1. The use of CI to characterize the amount of homoplasy for a particular tree is 
controversial, and other indices have been suggested, including the homoplasy excess ratio 
(Archie, 1989) and the retention index (Farris 1989b). See also papers by Farris ( 1990) and 
Archie (1990) for various opinions regarding these measures and the use of CI. 

2. Nelson (1979) is often credited as the source for strict consensus trees. However, he 
really described component analysis, which possesses qualities of clique analysis (Page, 
1987, 1988, 1989). The resulting trees may fit the description of either strict or majority rule 
consensus trees or no particular consensus tree at all, depending on the number of trees 
involved. The use of consensus trees of all types has increased in the last few years, probably 
because they offer what appears to be a simple solution to the difficult problems associated 
with choosing among several equally parsimonious trees. However, we feel that consensus 
trees solve no such problems and their use in this manner has the effect of avoiding the 
difficult problems associated with equally parsimonious trees. Having many equally 
parsimonious trees may be the result of insufficient study (a lack of sufficient data) or 
chaotic evolution in the particular data sets studied, reflecting a complex history of 
reticulation or homoplastic evolution. You should be extremely careful when using 
consensus trees to investigate specific questions or to organize your data. A consensus tree 



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KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



should not be presented as a phylogeny unless it is topologically identical with one or more 
of the most parsimonious trees. 

3. Competing trees fall into three overlapping categories. 1 ) The competing trees may be 
equally parsimonious. 2) They may be equally parsimonious but have different topologies. 
3) They may be close but not equal in terms of tree length. Equally parsimonious trees that 
have the same topology result from different interpretations of character evolution. Fre- 
quently, this involves equally parsimonious interpretations of parallelism and reversals. 
When different topologies are involved, the trees have different knowledge claims concern- 
ing the relationships of the taxa studied. Such trees may be logically consistent or logically 
inconsistent with each other. You can use the Adams consensus method to explore these 
differences. Loss of resolution occurs when there is an increase in the number of nodes with 
polytomies or when the number of branches involved in a polytomy increases. The taxon 
that causes a loss of resolution must have different placements on at least two trees. You can 
check the original trees just in these areas, or you can do pairwise comparisons of the 
original trees to identify the conflicts. Once these trees have been identified, you can find the 
taxa involved and the characters that are responsible for the conflict. You might also use the 
strict consensus method to determine what monophyletic groups are supported by all of the 
original trees. Finally, you might use the Adams method for a set of trees that are within 
certain tree-length values of each other. Funk ( 1985) found this approach useful in her study 
of hybrid species, when the differing placements of certain species helped spot potential 
hybrids and lead to hypotheses concerning their parental species. Table 5.7 summarizes 
some of the uses and potential problems with various kinds of consensus trees. 

Table 5.7. — Uses and potential problems of the various kinds of consensus trees. 



Kind of tree 



Questions asked 



Characteristics 



Strict 



Adams 



Majority 



1 ) What groups are always 
monophyletic? 



1) What is the most highly 
resolved tree that will identify 
problem taxa? 

2) Are these trees logically 
consistent? 

1 ) What is a summary of the 
competing trees where the 
dominant pattern prevails? 



1 ) Loss of resolution may be extreme. 

2) Useful as a phylogeny only if 
topologically identical with one of 
the original parsimony trees. 

1 ) "Strange" taxon placements not 
found in any original tree. 

2) Useful as a phylogeny only if 
topologically identical with one of 
the original parsimony trees. 

1 ) Most useful when there are very 
little conflicting data. 

2) Useful as a phylogeny only if 
topologically identical with one of 
the original parsimony trees. 

3) Ties can be avoided only if you 
start out with an odd number of 
trees (A. Kluge, pers. comm.). 



Chapter 6 
CLASSIFICATION 



A classification is a groups-within-groups organization of taxa of organisms. This type of 
organization can be represented in many different ways. Most classifications take the form 
of Linnaean hierarchies, with the relative positions of groups and subgroups being tagged 
with a Linnaean rank (phylum, family, genus, etc.). Classifications can also be represented 
graphically, using a tree structure or a Venn diagram (see Chapter 1 ). 

Nothing marks the phylogenetic system as different from competing systems so much as 
the issue of classification. Many investigators are drawn to the advantages of using phyloge- 
netic techniques to infer common ancestry relationships, but they seem to balk at excluding 
paraphyletic groups. Much acrimony is produced in the name of "tradition," "common 
sense," or other emotional side issues, rather than focusing in on the one central issue: 
should biological classifications be based on phylogeny? We answer "yes" because we are 
evolutionary biologists. Actually, phylogenetic systematists operate under only two basic 
principles. First, a classification must be consistent with the phylogeny on which it is based. 
Second, a classification should be fully informative regarding the common ancestry rela- 
tionships of the groups classified. Further, we recommend that while implementing these 
principles, you make every attempt to alter the current classification as little as possible. 

The principle of consistency can be met if the investigator includes only monophyletic 
groups. The principle of information content can be approached by several means. The 
conventions we outline in this chapter will allow you to change existing classifications 
minimally to bring them "in line" with current hypotheses of genealogical relationships. The 
principle of minimum change represents a conservafive approach that provides for historical 
continuity of classification in the change from preevolutionary concepts (such as distinctive- 
ness and identity) to evolutionary concepts (common ancestry relationships). 

Often, taxonomists are categorized as either "splitters" or "lumpers." Actually, 
phylogenedcists are neither. By following the principles of monophyly and maximum 
information content, we seek to establish classifications that reflect natural groups. Some- 
times this requires breaking up a paraphyletic or polyphyletic group into smaller groups. At 
other times these goals are met by combining smaller groups into more inclusive groups. 
The categories of "splitters" and "lumpers" belong to the past when authority was more 
important than data. 

The topics covered in this chapter are 1 ) how to evaluate existing classificafions relative to 
new ideas about the common ancestry relationships of the organisms studied and 2) how to 
constmct classificafions using conventions designed to conserve as much of the old taxono- 
mies as possible. 



92 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

Evaluation of Existing CLASSincATiONs 

Once the investigator has analyzed her data and arrived at a phylogenetic tree (a 
hypothesis of common ancestry relationships), she will wish to compare her results with the 
ideas found in previous studies. Most of the ideas of relationship that exist in the literature 
are embodied in classifications. These classifications have connotations regarding the 
relationships that the previous investigator felt were important to the taxonomy of the group. 
The ideas embodied in the classification may represent the intuition of the previous 
investigator or they may refer very specifically to some evolutionary principle the investiga- 
tor had in mind. Whatever the ideas held by the original taxonomist, it is up to our 
investigator to compare her hypothesis with the existing classification to determine how that 
classification should be changed to bring it in line with the phylogeny. This is accomplished 
by comparing the structure of the classification with the structure of the phylogenetic tree. If 
the exisfing taxonomy is logically consistent with the phylogeny, then the first criterion is 
met. If it is fully informative about the phylogeny, then the second criterion is met. Because 
logical consistency is the most basic requirement, we will cover it first. 

Logical Consistency 

As outlined by Hull (1964), logical consistency exists between a classification and a 
phylogeny if, and only if, at least one phylogeny can be derived from the classification that 
is, itself, the phylogeny. You might get the idea that it may be possible to derive more than 
one phylogeny from a classification. This is true for some classifications but not for others. 
However, a more basic question must be addressed. How do we actually go about compar- 
ing a phylogeny and a classification? The answer is fairly straightforward. The phylogeny 
exists in tree fomi and the classification exists in the form of a groups-within-groups 
hierarchy. We therefore must convert the hierarchy of the classification into tree form. Once 
in tree form, we can make a side-by-side comparison. 

Example 6.1. — Converting a classificafion to tree form. 

The following classification is of the family Goodeidae. North American and Middle 
American killifishes related to the guppies and swordtails and commonly found in aquarium 
stores (Paremi, 1981). 

Family Goodeidae 

Subfamily Empetrichthyinae 

Genus Empetrkhthys 

Genus Crenichthys 
Subfamily Goodeinae 

(several genera) 

We convert the classification to tree form using the groupings inherent in the classification; 
for example, there are two subfamilies within the family. The classification in tree form 



CLASSIFICATION 93 

looks like Fig. 6. 1 . (We have included the names of the groups and subgroups to show the 
direct connection between this tree and the original classification. It is not necessary to do so 
once you have the idea firmly in mind.) 



Empetrichthys Crenichthys Goodeinae 




/ 

Empetrichyinae^ 



Goodeidae 
Fig. 6.1. — Parenti's (1981) classification of Goodeidae in tree form. 



There are several important things about this classification in tree form. First, it exactly 
reflects the groups-within-groups organization of the ranked classification. Intemodes take 
the place of higher group names. Second, it is entirely dichotomous. This means that no 
additional diagrams can be derived from it. Classifications that are fully ranked and thus 
have tree structures that are entirely dichotomous have no derivatives. Last, it is not a 
phylogeny. We cannot stress this last point enough. The classification might be based on 
phenetic principles. It might be arbitrary. It might be intuitive. Or, it might be phylogenetic. 
To know which, we would have to read the paper. None of us is an expert in goodeid 
relationships, but it turns out that Parenti (1981 ) also performed a phylogenetic analysis, so 
we can take her hypothesis as a basis for comparison (Fig. 6.2). 

Now, we can place the classification beside the phylogeny. Note that they are entirely 
identical in their topologies. No claims of group-within-group relationships are different 
from the claims of common ancestry. Let us imagine that Parenti had chosen to classify her 
groups in a different way. 

Family Goodeidae 

Subfamily Empetrichthytnae 

Genus Empetrichthys 
Subfamily Crenichthyinae 

Genus Crenichthys 
Subfamily Goodeinae 

(several genera) 



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KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



Empetrichthys Crenichthys Goodeinae 




No pelvic fins or fin supports -- 
"Y"-shaped first epibranchial -- 




-- Viviparous reproduction 

-- Unbranched anterior anal fin rays 

First anal fin ray rudimentary 

-- Muscular urogenital system in 
males 

-- Other characters associated with 
viviparous reproduction 



-- First 2-7 middle anal radials absent 
or fused to proximal radials 

-- Distal arm of premaxilla straight 
-- Dorsal processes of maxilla reduced 



Fig. 6.2. — A phylogenetic tree of goodeid relationships (from Parenti, 1981). 

Conversion of the classification to tree form would produce Fig. 6.3. 

We now have a situation in which there is a trichotomy among the three subgroups. 
Whenever there is a polytomy, we can derive additional hypotheses about possible 
subgroupings. These hypotheses represent all of the possible resolutions of the classification 
tree, given its structure. That is, we can derive possible resolutions that are consistent with 
the original classification. Following the principle of logical consistency, if one of these 



Empetrichthys Crenichthys Goodeinae 




Goodeidae 



Fig. 6.3. — An alternative classification of goodeids in tree form. 



CLASSIFICATION 



95 



derivatives is the phylogeny, then the classification is logically consistent with the phylog- 
eny even if it is not fully informative about the relationships hypothesized in the phylogeny. 
Tlie number of possible trees can be determined mathematically. For the trichotomy in Fig. 
6.3, there are three dichotomous resolutions (Fig. 6.4a-c). Note that one of them (Fig. 6.4b) 
is topologically identical to the phylogeny. Therefore, even though the branching topology 
of the classification (Fig. 6.3) is not the same as the topology of the phylogenetic hypothesis 
(Fig. 6.2), the classification is logically consistent with the phylogeny because one possible 
derivation of the classification is topographically identical to the phylogeny. 







<^^^ 



.N^^^ 



/ 



..^^ 



r-^^ 



.K^^^ 



0^' 



.<^^^ 



^2^ 



,^ 



^^ 



O^ 





0^^ 



.^.^^ 



kV^ g"^^^ ^^ 

r.^ J ^6^ 



<c> 



^^ 



(^ 



Goodeidae 




Goodeidae 



Fig. 6.4. — ^Three possible phylogenetic trees of goodeids derived from the relationships implied by the 
classification in Fig. 6.3. 



Working out derivative classifications in tree form is easy when the number of possible 
derivatives is small. The method becomes cumbersome, however, when the number of 
possible derivative trees is large. Fortunately, there is an alternative method for checking the 
logical consistency of a classification with a particular phylogeny, the use of Venn diagrams. 



96 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



In Fig. 6.5, we have converted the phylogeny (Fig. 6.5a), the first classification (Fig. 6.5b), 
and the second classification (Fig. 6.5c) into Venn diagrams. We then "layer" all three 
diagrams and check for overlap between ellipses, which is equivalent to checking for 
overlapping sets in set theoiy. If there are no overlaps, the classification(s) is logically 
consistent; if there is overlap, the classification(s) is logically inconsistent. In this case there 
is no overlap, so both classifications are logically consistent with the phylogeny (Fig. 6.5d). 




Goodeinae 



[Empetnchthys) ( Crenichthys j 

Empetrichyinae 



f Goodeinae j 



Goodeidae 



Empetnchthys) f Crenichthys j f Goodeinae j 

Goodeidae 



Empetnchthys 




Crenichthys 



Goodeinae 




Fig. 6.5. — Venn diagrams of goodeids. a. The goodeid phylogeny. b, c. Two different classifications. 
d. The result of combining a, b, and c into a single diagram. 



CLASSIFICATION 97 

You can easily see that a classification can be logically consistent with a phylogeny 
without being fully informative about the common ancestry relationships implied in the 
phylogeny. In fact, the only way to have a classification that is logically inconsistent with the 
phylogeny is to have a classification for which we can derive no tree that is the original 
phylogeny. More formally, a classification is logically inconsistent with a phylogeny if no 
derivative of that classification is the original phylogeny. In terms of our Venn diagrams, a 
classification is logically inconsistent with the phylogeny if its Venn diagrams have overlap- 
ping ellipses. In other words, they are logically inconsistent if they violate the inclusion/ 
exclusion rule. 

We may be making it sound like most classifications are logically consistent with the 
phylogeny you are likely to generate with your analysis. Nothing could be further from the 
truth. You will find that most classifications are logically inconsistent with your hypotheses 
of common ancestry. Why? Because most existing classifications contain paraphyletic and 
even polyphyletic groups. Let's look at the effects of the inclusion of a paraphyletic group in 
the correspondence of classifications to phylogenies. 

Example 6.2. — The very distinctive Ciis. 

Investigator Smith has performed a phylogenetic analysis on the genera comprising the 
family Cidae. She has arrived at the phylogenetic hypothesis shown in Fig. 6.6. This family 
was well known to previous investigators. What struck these investigators was how different 
members of the genus Cm were from other members of the family. This distinctiveness was 
embodied in the traditional classification. 

Family Cidae 
Subfamily Ainae 

Genus Aus 

Genus Bus 
Subfamily Cinae 

Genus Cus 

Smith wants to know if the current classificadon is logically consistent with her phyloge- 
netic hypothesis. To do so, she must perform the following steps. 

1 . She prepares a Venn diagram for the classification and another Venn diagram for the 
phylogeny. 

2. She layers the Venn diagram of the classification over the Venn diagram of the 
phylogeny. 

3. If ellipses do not overlap, then she knows that the classification is logically consistent 
with the phylogeny. If one or more ellipses overlap, then the classification is logically 
inconsistent with the phylogeny. 



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KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. ly 



Aus Bus 



Cus 




Fig. 6.6. — The phylogeny of the family Cidae (Example 6.2). The phylogeny and the relative amounts of 
change along each branch are taken as "true." 



Therefore, Smith does the following for the Cidae. 

1. The classification is converted to a classification tree (Fig. 6.7a) and then into a Venn 
diagram (Fig. 6.7c). (As you gain experience, you can draw the Venn diagram directly from 
the classification.) The phylogeny (Fig. 6.7b) is then converted into a Venn diagram (Fig. 6.7d). 



Aus 



Bus 



Cus 




ABC 



Aus 



Bus 



Cus 




ABC 



c 



Aus Bus 



J GD 



GiD C 



Bus Cus 



) 



c 



Aus 







Bus ) Cus 



Fig. 6.7. — The Cidae classification, a. Tree form. b. The phylogeny. c. Venn diagram of a. d. Venn diagram 
of b. e. Venn diagram combining c and d. 



CLASSIFICATION 99 

2. The Venn diagrams are layered (Fig. 6.7e). 

3. The Venn diagram of the phylogeny (Fig. 6.7d) and the Venn diagram of the classifica- 
tion (Fig. 6.7c) overlap (Fig. 6.7e); Bus is a member of the group BC in the Venn diagram of 
the phylogeny, and it is a member of the group AB in the Venn diagram of the classification. 
Therefore, the classification is logically inconsistent with the phylogeny. 

Smith, realizing that she cannot tolerate a classification that is logically inconsistent with 
the evolution of the Cidae, creates the following classification. 

Family Cidae 
Subfamily Ainae 

Genus Aus 
Subfamily Cinae 

Genus Bus 

Genus Cus 

Determining the Number of Derivative Classifications 

Although Venn diagrams are the most direct route to determining whether a classification 
is logically consistent with a phylogeny, you might also wish to calculate the number of 
derivative classification trees for a particular classification. The number of alternative 
classifications that can be derived from a particular classification is directly related to the 
number of polytomies in the classification's branching structure. Felsenstein ( 1 977) presents 
tables to detemiine the number of tree topologies that can be derived from a basic tree with 
multifurcations (polytomies). We have reproduced parts of one of these tables as Table 6.1. 
Note that the numbers refer only to the number of terminal taxa. Other tables must be 
consulted if ancestors are included. 

Example 6.3. — Classification of the hypothetical Xaceae. 

The Xaceae is classified into three major subgroups, as shown in the classification tree in 



Table 6. 1 . — The total number of possible derivative trees for polytomies of n branches. Intemodes cannot 
be occupied by "ancestors'" (from Felsenstein, 1977). 

n All trees Dichotomous trees 

3 4 3 

4 26 15 

5 236 105 

6 2752 945 

7 39,208 10.395 

8 660,032 135,135 

9 12,818,912 2,027.025 
10 282,137,824 34,459,425 



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KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No, 19 



Xus 



Yus 



Zus 




Fig, 6.8. — The classification tree of the Xaceae (Example 6.3). 



Fig. 6.8. We calculate the number of trees that can be derived from this classification in the 
following manner. 

1 . Determine if the phylogeny to be compared is dichotomous or contains polytomies. In 
this example, we assume that it is a dichotomous phylogeny. 

2. Select the appropriate column in Table 6. 1. In this case, we use the column on the right 
(dichotomous trees). 

3. Determine the number of branches for each polytomy. 

4. Using Table 6. 1 , find the number of trees possible for each polytomy. Multiply all of 
the values obtained in step 3 together to obtain the total number of derivative classification 
trees: 

(3)(3)(15)= 135. 

Thus, there are 1 35 possible dichotomous trees that can be derived from our classification of 
Xaceae. 



Classification Evaluation Exercises 

Each exercise consists of one or more classifications and a phylogenetic tree. You are 
asked to 1) convert the classification into tree form, 2) state the number of possible 
derivative trees that can be obtained, and 3) evaluate the classification in temis of its 
consistency with the phylogeny. 



EXERCISE 6.1. — Phylogeny of the Recent tetrapod vertebrates. 

For classifications, see Fig. 6.9. For the phylogenetic tree, see Fig. 6.10. 



CLASSIFICATION 



101 



Lissamphibia 

Reptilia 
Chelonia 
Lepidosauria 
Crocodylia 

Aves 

Mammalia 



Lissamphibia 

Mammalia 

Chelonia 

Lepidosauria 

Crocodylia 

Aves 



C Tetrapoda 

Lissamphibia 
Amniota 
Mammalia 
Reptilia 
Chelonia 
Sauria 

Lepidosauria 
Archosauria 
Crocodylia 
Aves 

Fig. 6.9. — Three classifications of Recent tetrapod vertebrates (Exercise 6. 1 ). 



.6A^ 



^^^ 



.<^' 



^>^ 



9- 



Feathers -- 



.x®' 



:.6° 



# 



O^ 



J" 



Long hard palate 

Pneumatic spaces in the -- 
skull 



Hooked 5th metatarsal 



Diapsid skull 



"Right-handed" circulatory system 



\0 



,*'> 



AmnJote egg 



# 



.<^ 



.<^ 



&■ 



■-O 



# 







Pedicellate teeth 



Tetrapod limbs 



Fig. 6.10. — A phylogenetic hypothesis of Recent tetrapod relationships (Exercise 6.1). 



102 KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 

EXERCISE 6.2.— Phylogeny of the land plants. 
Use the following classification. 

Division Bryophyta 

Class Anthoceropsida 

Class Marchantiopsida 

Class Bryopsida 
Division Tracheophyta 
Subdivision Psilotophytina 
Subdivision Lycopodophytina 
Subdivision Sphenophytina 
Subdivision Pteridophytina 
Subdivision Spermatophytina 

Class Cycadopsida 

Class Pinopsida 

Class Ginkgopsida 

Class Gnetopsida 

Class Angiospennopsida 

For the phylogenetic tree, see Fig. 6. 11 . 

Constructing Phylogenetic Classihcations 

There are two basic ways to construct phylogenetic classifications. First, one can consis- 
tently place sister groups in the classification at the same rank or rank equivalent. In such a 
classification, every hypothesized monophyletic group is named. If this manner of classify- 
ing is adopted, rank within a restricted part of the classification denotes relative time of 
origin. Second, one can adopt a set of conventions designed to reflect the branching 
sequence exactly but not require that every monophyletic group be named. 

We consider it beyond the scope of this workbook to detail the controversies surrounding 
such topics as the suitability of Linnaean ranks versus indentation or numerical prefixes for 
constructing classifications (reviewed in Wiley, 1 98 1 a). Instead, we will briefly review some 
of the basics of phylogenetic classification and provide a summaiy of some of the conven- 
tions you might use in constructing your classifications. This will be followed by some 
exercises designed to demonstrate when certain conventions might be used. 

Rules of Phylogenetic Classifications 

Rule 1. — Only monophyletic groups will be formally classified. 

Rule 2. — All classifications will be logically consistent with the phylogenetic hypothesis 

accepted by the investigator. 
Rule 3. — Regardless of the conventions used, each classification must be capable of 

expressing the sister gi'oup relationships among the taxa classified. 



CLASSIFICATION 



103 



8^ x6* 






^"^^ .<>^ ..<<^ 



.^" -i<>^ ^' .«^^ 



>&■ 






.# 



6°"^ ^6°^ -'^^ 



.6^ 



3 c:P 



,^" 



Spermatophytina 



Tracheophyta 



Fig. 6.1 1. — A phylogenetic hypothesis of selected land plant relationships (Exercise 6.2). (Name endings 
for terminal taxa correspond to classification for convenience.) 



Conventions 

We list the conventions used by Wiley (1981a) in his "annotated Linnaean hierarchy" 
system. 

Convention 1. — The Linnaean system of ranks will be used. 

This is one of the controversial conventions. There have been many attempts to substitute 
other means of subordinating taxa. and some of these attempts are mentioned at the end of 
this chapter. None are, in our opinion, satisfactory. However, the Linnaean system of ranks 
is, itself, a convention. There is no biological or scientific imperative for using Linnaean 
ranks. 



Convention 2. — Minimum taxonomic decisions will be made to construct a classification 
or to modify existing classifications. 



104 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

This convention can be met when making a new classification by rejecting redundant 
taxon names. For example, if a family has only one genus, do not coin a subfamily name. 
The subfamily name would be redundant with the family name. Of course, the family name 
is redundant, in temis of its diagnosis, with the genus. But, the family name serves another 
purpose, to place the genus within the context of its phylogenetic relationships with other 
genera. When converting existing classifications, make every effort to retain well known 
taxon names at their traditional ranks. This may or may not be possible. 

Convention 3. — Taxa forming an asymmetrical part of a phylogenetic tree may be placed at 
the same rank and sequenced in their order of branching (Nelson, 1 972). When such a list 
is encountered, the sequence of the list denotes the sequence of the branching. 

This convention can be used by an investigator who decides that he will not name every 
monophyletic group but still wishes to preserve Rule 3. This convention is frequently 
termed the "listing convention" or "sequencing convention" and was first proposed by 
Nelson (1972) (Fig. 6.12). 



Myxiniodea Petromyzontia Gnathostomata 




Subphylum Vertebrata 
Infraphylum Myxiniodea 
Infraphylum Petromyzontia 
Infraphylum Gnathostomata 

Fig. 6. 1 2. — A hypothesis of the relationships and a classification of three vertebrate groups, illustrating the 
sequencing convention (Convention 3). 



Convention 4. — Taxa whose relationships are polytomous will be placed sedis miitahilis at 
the same rank (Wiley, 1979). 

This convention must be used if the sequence convention is used because you must be 
able to discriminate a list that translates into a series of dichotomies from a list that translates 
into a polytomy (Fig. 6.13). 

Convention 5. — Monophyletic taxa of uncertain relationships will be placed incertae sedis 
at a level in the hierarchy where their relationships ai'e known with some certainty. 

Tliis convention covers the situation where a small inonophyletic group is thought to 
belong to, say, an order but cannot be placed in any suborder, family, etc., within that order. 



CLASSIFICATION 



105 






r ^i?>> sr^ .^^ 







Subphylum Vertebrata 

Infraphylum Myxiniodea (sedis mutabilis) 
Infraphylum Petromyzontia (sedis mutabilis) 
Infraphylum Gnathostomata [sedis mutabilis) 

Superclass Chondrichthys 

Superclass Teleostomi 

Fig. 6. 1 3. — Another hypothesis of the relationships among vertebrates, with gnathostomes resolved into 
its two major subgroups, illustrating the sedis miitahilis convention (Convention 4). 



Convention 6. — A group whose qualities are not known may be included in a phylogenetic 
classification if it is treated as incertae sedis and its name is put in shutter quotes 
(quotation marks) (Wiley, 1979). 

One of the common problems you will encounter is the leftover taxa (candidates for 
incertae sedis under Convention 5) and "taxa" that are peripheral to your problem and are 
neither demonstrably monophyletic, paraphyletic, or (rarely) polyphyletic. Sometimes, 
these groups are well known. To omit them would leave the classification incomplete and 
make a statement that you know they are not monophyletic; to put them in without 
qualification would imply that you consider them monophyletic. The shutter quotes carry 
the connotation that all included subtaxa are presently incertae sedis at the place in the 
hierarchy where you put the "taxon." This convention should be used with caution. For 
example, it is not advisable to treat a polytomy occurring in the middle of a phylogenetic 
hypothesis (see Fig. 6.14) with this convention. 



Aus Bus 



Dus Eus 




Fig. 6.14. — Relationships among a hypothetical group of genera, illustrating the unwise use of shutter 
quotes (Convention 6). 



106 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

Convention 7. — Fossil taxa will be treated differently than Recent taxa. Fossil taxa will 
always be sequenced with their Recent relatives following Convention 3. If they are 
ranked, their status as fossils will be denoted by placing a dagger or cross symbol before 
the rank (Nelson, 1972). Alternatively, they may be given the neutral rank of "plesion" 
(Patterson and Rosen, 1977). As natural taxa, monophyletic fossil taxa may stand 
incertae sedis or sedis mutabilis, just as any Recent natural taxon. 

Wiley's original conventions (Wiley. 1979, 1981a) incorporated only the plesion conven- 
tion. The convention as stated here is less restrictive, but still keeps our ever-shifting 
understanding of the relationships of fossil groups from continually changing the hierarchi- 
cal classifications of Recent groups. Figure 6.15 illustrates two ways of classifying some 
fossil mammals with their Recent relatives. 



a Infradivision Theria 

Supercohort Marsupialia 
Supercohort Eutheria 



b Infradivision Theria C Infradivision Theria 

Plesion Kueneotherium tSupercohort Kueneotheria 

Plesion Symmetrodonta tSupercohort Symmetrodontia 

Plesion Dryolestoidea tSupercohort Dryolestia 

Plesion Paramus tSupercohort Paramia 

Supercohort Marsupialia Supercohort Marsupialia 

Supercohort Eutheria Supercohort Eutheria 

Fig. 6.15. — TTiree classifications of some mammals (see Convention 7). a. Classification containing only 
groups with Recent species, b. "Major" fossil groups added to the Recent classification using the plesion 
convention, c. The same groups added but using the listing and dagger conventions. 



Convention 8. — Stem species (ancestral species) are placed in classifications in parentheses 
beside the names of taxa they gave rise to or taxa containing their descendants, as 
appropriate. A stem species can be inserted into the hierarchy in one of three ways. 1 ) A 
stem species of a suprageneric taxon will be placed in a monotypic genus and inserted in 
the hierarchy beside the name of the taxon that contains its descendants. 2) A stem species 
of a genus will be placed in that genus and inserted beside the genus name. 3) The stem 
species of a species within a genus will be placed in that genus and inserted beside the 
species name. 

Stem species, as Hennig ( 1966) recognized, are equivalent to the supraspecific taxa that 
contain their descendants. This convention treats stem species in exactly this manner. We 
illustrate this convention with the hypothetical ancestor, recently "discovered," of the 
Sarcopterygii (Fig. 6.16). 



CLASSIFICATION 



107 



;^<>' 






;s<>" 



V^ 



>^- >^- ..<> 



O^^ 



o^° 







Superclass Teleostomi 
Class Actinopterygii 

Class Sarcopterygii {Sarcopterygius primus) 
Subclass Actinistia 
Subclass Dipnoiformes 
Infraclass Dipnoi 
Infraclass Choanata 



Fig. 6. 1 6. — A phylogeny of teleostome vertebrates and a classification illustrating the ancestor convention 

(Convention 8). 

Convention 9. — A taxon of hybrid origin will be indicated by placing the names of its 
parental species in parentheses beside the taxon 's name. If the taxon is placed beside one 
of the parental taxa, its sequence in a list carries no connotation of branching sequence 
relative to taxa of "normal" origin. 

Wiley (1981a:226-227) suggested that hybrids between species in different genera 
should be placed in a third genus. This is not necessary, but you should not place the hybrids 
in both genera if you do not wish to name a third genus because it might cause confusion 
regarding the actual number of species that exist. There are several ways this convention can 
be used. One way is shown in Fig, 6.17. You might also tiy listing the hybrid taxa under, say, 
the genus name and then proceed with the species of nonhybrid origin using other conventions. 



/./ 



^- ^^- ^- '^^^ '^^ \^ 



0^' . / 




Genus Mus 
M. kus 
M. lus 

M. mus {M. lus X M. nus) 
M. nus 
M. ous 
M. pus 



Fig. 6. 1 7. — A phylogeny of the hypothetical|genus Mus, with a classification illustrating the hybrid taxon 
convention (Convention 9). fpEtif 5 ci» •r^\' 



tiSf ^IddiTA-] 



108 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

The following classification of some species of the lizard genus Cnemidophorus shows a 
really nasty case handled in this manner. 

Genus Cnemidophorus 
species of hybrid origin: 

C. uniparens (inomatiis x gidaris x inornatus) 
C. neomexicamis {inornatus x tigris) 
C. laredoensis (se.xlineatus x gularis) 
C. '"tesselatus'' A-B Ctesselatus^' C-E x sexlineatus) 
C. ""tesselatus''' C-E (tigris x septemvittatus) 
C. tigris species group 

C. tigris 
C. sexlineatus species group 
sexlineatus squadron 
C. sexlineatus 
C. inornatus 
gularis squadron 
C. gularis 
C. septemvittatus 

This convention also works for taxa of symbiotic origin, except that the usual "x" sign 
that indicates hybrids is replaced by a "+" sign to indicate the "additive" nature of the 
resulting organism. 



Quick Quiz — Taxonomy vs. Systematics 

The Central Names Committee of the National Mugwort Society announces that it has accepted 
your recent phylogenetic study of spiny-toed mugworts. except that all of your higher order 
categories have been collapsed to subgenera. What should you do? 



Convention Exercises 

The following exercises are designed to give you some practice in using the conventions 
outlined above. These exercises are rather different from the exercises in the previous 
chapters because of the nature of the rules and the fact that ranks are relative. Also, note that 
the names of taxa are letters. You may use the letters as is, or if your instructor wishes, use 
them as the root to which a correct ending is added. 

EXERCISE 6.3. — From the phylogenetic tree in Fig. 6.18, do the following: 

1 . Use the sequence convention at every opportunity to produce a classification begin- 
ning at the hierarchical rank of order. For higher group names, use the following convention. 



CLASSIFICATION 



109 



BCDEFMNOPQR 




Fig. 6.18. — Aphylogeny of the hypothetical order A-S (Exercise 6.3). 

For a group composed of A, B, C, and D, use A-D; for a group composed of A and B. use 
AB, etc. 

2. Produce a second classification without using the sequencing convention, that is, by 
naming every branch point. For example, A-F is an order with two suborders, F and A-E. 
Don't forget to rank all sister groups at the same rank. 

3. Count the number of hierarchical levels you save by using the sequencing convention. 

EXERCISE 6.4. — From the phylogenetic tree of the hypothetical order A-U in Fig. 6.19, 
use Conventions 3, 4, and 5 to classify the members of the group using order, suborder, 
family, subfamily, and genus. 



BCD 



F 



M 



Q R S T U 




Fig. 6. 1 9. — A phylogeny of the hypothetical order A-U (Exercise 6.4). The terminal taxa are genera. 



110 



KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 



EXERCISE 6.5. — From the phylogenetic tree in Fig. 6.20, do the following: 

1. Naming every branch point, classify the fossil taxa without using Convention 7. 

2. Classify the Recent taxa first and then add the fossil taxa into the classification using 
Convention 7. 

3. Briefly compare the two results in terms of hierarchical levels used and the magnitude 
of change in the classifications brought about by shifting the relationships of one fossil 
species. 




Fig. 6.20. — A phylogenetic hypothesis of the relationships among genera of the hypothetical order 
Tusiformes, including Recent and fossil groups (Exercise 6.5). 



EXERCISE 6.6. — The very distinctive flightless cormorant. 

The Galapagos Cormorant is very distinctive in that unlike the rest of the members of the 
family Phalacrocoracidae, it has small nonfunctional wings, peculiai" wing feathers, and a 
massive bill. It was considered so distinctive that it was placed in its own genus, 
Nannopterum (literally "tiny wings'"). 

1 . Derive a phylogenetic tree from Table 6.2. The characters were polarized using the 
method of Madison et al. (1984) and the OG vector. 

2. Evaluate the traditional classification below in terms of its consistency with the 
phylogeny. 

Phalacwcorax africanus 

P. pygmaem 

P. penicillatus 

P. carbo 

P. awitus 

P. olivaceus 
Nannopterum harrisi 



CLASSIFICATION 



111 



Table 6.2. — Data matrix for cormorant genera Phalacrocorax and Naimoptcntm (Exercise 6.6) (data from 
Siegel-Causey, 1988). 











Transformation 


series 








Taxon 


8 


15 


17 


30 


51 


80 


82 


86 


93 


OG 





























P. afiicaniis 


1 


























P. awitiis 














1 




1 





1 


P. carho 











1 













1 


N. harrisi 





1 
















1 





P. olivaceus 














1 










I 


P. penicillatus 






















1 





P. pygmaeus 


1 





1 





















3. Construct a new classification if necessary. 

Chapter Notes and References 

1 . The criterion of logical consistency has also been discussed by Wiley ( 198 lb, 1987a, 
1989). 

2. The literature on biological classification and its relationship to evolution is huge. The 
texts mentioned in Chapter 1. as well as the pages of Systematic Zoology, provide an 
introduction. 

3. For an excellent introduction to some problems perceived in using the Linnaean 
system in phylogenetic classifications and a review of some of the proposed solutions, see 
Griffiths ( 1976). There are three basic options for adopting a set of conventions similar to 
that outlined here. First, one could simply use an unannotated Linnaean system of categories 
and name every branch. Second, one could adopt a numbering sequence (Hennig, 1981; 
L0vtrup, 1973). Third, one could abandon both numbering and prefixes and classify by pure 
indentation (cf. Gauthier et al., 1988). 



Quick Quiz Answers 
Taxonomy vs. Systematics 

Relax. It's the relationships that count for the systematist, not the names. Anyway, science by committee 
is for administrators and the faint hearted, not for you. Continue on as before. 



112 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



Chapter 7 

COEVOLUTIONARY STUDIES 



Phylogenetic trees can be used in a variety of ways in evolutionary biology. Many of these 
applications involve comparing the degree to which the history of one group coincides with 
the history of the geographic areas in which its members reside or with the histories of other 
groups. Brooks (1985) suggested two basic reasons why a species might live where it lives 
or be associated with the particular species it is associated with. A species may live in a 
certain geographic area because its ancestor lived in that area and the descendant evolved 
there. Altematively, the species may have evolved elsewhere and dispersed into the area 
where it now resides. In the first case (association by descent), we would expect the history 
of the species to coincide with the history of the area; whereas in the second (association by 
colonization), we would not. Likewise, two or more species that exist in a close ecological 
association may be associated because their ancestors were associated, or they may be 
associated because they evolved in association with other species and subsequently 
"switched allegiances." In the first case, we would expect the histories of the taxa involved 
in the association to coincide (to be congruent): whereas in the second case, we would not 
expect to find such congruence. Taxa that show historical congruence either with geographic 
areas or with other taxa are said to exhibit cospeciation patterns. Phylogenetic systematic 
methods can help distinguish associations due to ancestral association from associations due 
to dispersal and colonization. 

In this chapter, we will learn how to use the results of phylogenetic analyses to study other 
aspects of the evolution of taxa. The major skills you will learn are 1 ) how to code the entire 
phylogenetic tree of a clade and 2) how to use the resulting matrix to study problems 
concerning biogeographic or coevolutionary aspects of the evolution of the clade or clades 
in the study. 

Coding Phylogenetic Trees 

Biogeographic and coevolutionary studies are concerned with the correlation of one set of 
data with another. For this correlation, the data can be partitioned into independent and 
dependent variables. For example, if we wish to study the relationships among a number of 
biogeographic areas based on the species that occur in these areas, we could consider the 
areas the independent variables and the phylogeny of the species the dependent variables. As 
another example, we might wish to assess the amount of cospeciation that has occurred 
during the evolution of a clade of parasites and their hosts. The relationships among the hosts 
would be considered the independent variables, and the relationships within the clade of 
parasites would be considered the dependent variables. The "fit" would be a function of the 
correlation between the host phylogeny (analyzed independently) and the parasite phylog- 



1 14 KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 1 9 

eny. Another way we could use coded trees is to arrive at a hypothesis about the relationships 
of areas or hosts. We can code a tree in the following manner. 

1. Label all nodes on the phylogenetic tree (Fig. 7.1). The tree will now consist of the 
branching structure, the terminal taxa, and labels for the ancestral nodes. 

A. aus A. bus A. cus 




Fig. 7. 1 . — The relationships among three hypothetical species, with labels (X, Y) for their ancestral nodes. 



2. Prepare a list of the occurrence of each terminal taxon relative to the independent 
variable. In this case the independent variables are areas of occurrence. 

Area 1 : Aus aus 
Area 2: Aus bus 
ATea3:Aus cus 

3. Begin constructing a matrix (Table 7.1). By convention, rows are the independent 
variables (areas), and columns are the dependent variables (taxa). Each row is composed of 
binary values that can be inferred from the original phylogenetic tree. In our case, we have 
three rows (areas) and tive columns (one for each branch and intemode). Assign to a cell the 
value " 1 " if either a taxon or its ancestor can be inferred to occur or have occurred in the area. 
In biogeographic studies, the assumption is that the ranges of ancestors are the additive 
ranges of their descendants. The same assumptions are used in coevolutionary studies; for 
example, the range of an ancestral parasite is the additive range of its descendants relative to 



Table 7. 1 . — Data matrix of the inferred distributions for taxa in Fig. 7.1. 









Taxa 






Area 


A. aus 


A. bus 


A. cus 


X 


Y 


1 


1 











1 


2 





1 





1 


1 


3 








1 


1 


1 



COEVOLUTIONARY STUDIES 115 

its hosts. In our simple example. Area 2 would be scored"!'" for the /4. /w5 and the X and Y 
columns, reflecting the observed occurrence of A. bus in Area 2 and the inferred occurrence 
ofX{A. bus's ancestor) and Y (the ancestor of the entire species group) in the same area. 
Let's take a closer look at this matrix. The first thing that should strike you is that there are 
three different sorts of data. 1 ) There are observational scores. Area 1 is scored " 1 " for A. aus 
as a matter of observation. 2) There are inferential scores. Area 1 has been scored "1" for 
hypothetical ancestor Y because, under the assumption of additive ranges, we can infer that 
the ancestor of the clade was in Area 1 based on the presence of one of its descendants {A. 
aus). 3) There are negative observational scores. Area 1 is scored "0" for the A. bus cell 
because A. bus was not observed in Area 1 . From this, we can infer that X was also absent 
from Area 1 . 

4. You now have a choice of methods, depending on what you know about the indepen- 
dent variables. If the relationships among the independent variables are not known, you can 
use the data to estimate these relationships. In this case, solve the tree for the dependent 
variable using the data from the independent variable. Do this just as if the coded tree were 
a matrix of characters for the areas. If an independent estimate of the independent variable is 
available, map the occurrence of taxa, including ancestors, on the tree that already exists for 
the areas. You can then use various tree statistics to judge the amount of coevolution or 
vicariance that can be inferred. 

When you think about it, it doesn't make much sense to solve the relationships of the 
dependent variable with the information based on a single clade. Why? Because a single 
clade is to an independent variable what a single transformation series is to a clade. 
Therefore, using only single clades. we are restricted to an 6f priori hypothesis of the 
relationships among the independent variables. Before we discuss methods for handling 
more than one clade, let's look at a real example involving only one group. 



Quick Quiz — Biogeography 

1 . What are the relationships of areas 1 , 2, and 3 impHed by the distributions of the three species 

0fA(/5? 

2. If you only used "presence and absence" data, what would you conclude about the relation- 
ships of these areas? 



Example 7.1. — Biogeography of the Amphilinidea. 

Amphilinids are a small group of parasitic flatworms found on several continents 
(Bandoni and Brooks, 1987). Their phylogenetic relationships are shown in Fig. 7.2. 



116 



KLl MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 



JAP 



FOL 



ELO 



LIG 



AFR 




Fig. 7.2. — The relationships among five species of parasitic flatworms (from Bandoni and Brooks, 1987) 
(Example 7.1 ). See text for abbreviations. 



1 . Label the nodes (already shown in Fig. 7.2). 

2. Prepare a list showing the occurrence of each of the terminal taxa. 

Amphilina japonica (JAP): North America (NA) 
A.foliacea (FOL): Eurasia (EU) 
Gigantolina elongate (ELO): Australia (AS) 
Schizochoerus Uguloideus (LIG): South America (SA) 
S. africamis (AFR): Africa (AF) 

3. Prepare a table of areas x taxa. Note that we have carefully rearranged the order of the 
taxa to put ancestral columns as close to relevant descendants as possible (Table 7.2). Tliis is 
not necessary, but it helps when you enter data and proof the matrix. 

4. We now have a choice. We can either solve the relationships of the areas or we can 
map the information in our matrix onto a previous hypothesis of area relationships. Or, we 
can do both and then compare the results. In this case, geologic evidence has been used to 
produce an "'area cladogram" showing the historical connections among the areas (Fig. 7.3). 



Table 7.2. — Data matrix of the inferred distributions of the Amphilinidea (Example 7.1). 













Taxat 










Area* 


JAP 


FOL 


u 


LIG 


AFR 


s 


ELO 


T 


V 


NA 


1 





1 



















EU 





1 


1 



















SA 











1 





1 





1 




AF 














1 


1 





1 




AS 




















1 


1 





* NA = North America. EU = Eurasia, SA = South America, AE = Africa. AS = Australia. 

t JAP = Amphilina japonica. FOL = A.foliacea. LIG = Schizochoerus lingiiloideus. AFR = S. africanus, ELO : 
Gigantolina elongata. S-V are ancestral taxa. 



COEVOLUTIONARY STUDIES 117 

NA EU AS SA AF 




Fig. 7.3. — The relationship.s among five continents based on geologic evidence and the relationships 
among amphilinid flatworms. See text for abbreviations. 



When we solve the relationships among the areas based on the area/taxon matrix, we find the 
same hypothesis. Thus, the relationships among areas as detemiined from geologic data and 
from amphilinid phylogeny are congruent. 

We may also estimate the level of congruence by mapping (fitting) the data for 
amphilinids directly onto the geologic hypothesis of the relationships among the continents. 
This approach treats the tree of continents as the independent variable. The matrix derived 
from the dependent variable (the continent x amphilinid matrix) could then be fitted to the 
hypothesis of continental relationships. This is easily accomplished in most computer 
programs such as PAUP, where we build a matrix, specify a particular tree (not necessarily 
derived from the matrix), and fit the matrix to the tree. We can then calculate an appropriate 
summary statistic. In this case, we select the CI, deleting the distributions of terminal taxa, 
and obtain a perfect fit (CI = 1.0). 

Note on single group analysis. Although this kind of analysis was originally designed to 
investigate biogeographic and coevolutionary problems, you might find it suitable for 
exploring other kinds of problems. For example, you and Dr. Fenetico have reconstructed 
different phytogenies for the saber-toothed cnidaria. You might use your tree as the 
independent variable and Dr. Fenetico's data as the dependent variables. One very pleasing 
result might be the discovery that his data fit your tree better than they do his tree. 

Single Tree Exercises 

In the exercises below, 1) code the tree, and 2) solve the relationships among the 
independent variables. Do not be confused by distributions and area labels. Each species 
occupies a single area and can be given a single label regardless of the number of drainages 
and states in which it occurs. We are simply presenting real examples. 



118 KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 

EXERCISE 7.1. — Luxilus zonatiis species group. 

This is a group of three species of North American minnows recently revised by Mayden 

(1988a). 

Luxilus cardinalis: lower Arkansas River drainage. Arkansas, Kansas, and Oklahoma, 

and Red River tributaries of the Ouachita highlands, Oklahoma (Area 1) 
L. pilsbiyi: White and Little Red rivers, Missouri and Arkansas (Area 2) 
L. zoiiatus: Ozark Plateau, north and east of L.pilsbryi in Missouri and Arkansas (Area 3) 

Luxilus cardinalis is the sister of L. pilsbiyi. and L. zoiiatus is the sister of this pair. 

EXERCISE 7.2. — Fuiululus iiottii species group. 

The Fundulus nottii species group consists of five North American topminnows with 
distributions shown below. (Fundulus blairae is actually more widespread than the informa- 
tion given, occurring along the northern Gulf Coast in sympatry with both F. nottii and 
eastern populations off. escambiae.) 

F. lineolatus (LEN): Florida peninsula and Atlantic Coast drainages (Area 1) 

F. escambiae (ESC): eastern Gulf Coast (Area 2) 

F. nottii (NOT): central Gulf Coast (Area 3) 

F. blairae (BLA): lower Mississippi River and western Gulf Coast (Area 4) 

F. dispar (DIS): upper Mississippi River (Area 5) 

Wiley (1977) hypothesized the following relationships for these species: 

((LIN(ESC, NOT))(BLA, DIS)). 



More Than One Group 

This method can also be used to compare the degree of congruence between geographic 
history and phylogeny for more than one group at a time. We treat each phylogenetic tree as 
a single chunk of the area x taxon matrix and perform a multigroup analysis. Analysis of 
multiple clades in the real world will almost certainly be more complicated than single clade 
analysis for two reasons. 1 ) Clades are not always found in all areas analyzed. For example, 
although family Aidae has members in all four of the areas you wish to study, family Bidae 
has members in only three of the areas. 2) Some clades may have members in two or more 
areas represented by single species in other clades. These are known as the "missing taxon" 
and "widespread species" problems. Both require us to consider character coding strategies 
that are not necessary when we perfomi single group analysis. Before attacking these 
problems, let us consider the simple example of two groups inhabiting the same region. We 
will use the classic hypothetical case presented by Humphries and Parenti (1986) and 
analyzed by Wiley (1987c, 1988a,b). 



COE VOLUTION ARY STUDIES 



119 



Example 7.2. — Two-group analysis. 

Consider the hypothetical example below for two taxa. Lizard (L) and Frog (F), inhabit- 
ing the same areas (from Humphries and Parenti, 1986). Tlie relationships of both groups are 
shown in Fig. 7.4. Their biogeographic ranges are Usted below. 

Li: Australia (AS) 

L2: New Guinea (NG) 

L3: South America (SA) 

L4: Africa (AF) 

F1:AS 

F2:NG 

F3: SA 

F4:AF 



L1 



L2 



L3 



L4 




F1 



F2 



F3 



F4 




Fig. 7.4. — Relationships and labeled ancestral nodes for a group of (a) lizards (L1-L4) and a group of (b) 
frogs (F1-F4) (from Humphries and Parenti, 1986) (Example 7.2). 



1. Label each ancestor on the tree of each group. (Fig. 7.4). 

2. Produce a binary coded matrix just as you would for a single group, but with both 
groups in the matrix (Table 7.3). 



Table 7.3. — Data matrix for lizards (L 1-L4) and frogs (F1-F4) and their ancestral taxa (x, y, z) (Example 7.2). 









- 








Taxa 














Area* 


LI 


L2 


Lx 


L3 


Ly 


L4 


Lz 


Fl 


F2 


Fx 


F3 


Fy 


F4 


Fz 


AS 


1 





1 





1 





1 


1 





1 





1 





1 


NG 





1 


1 





1 





1 





1 


1 





1 





1 


SA 











1 


1 





1 











1 


1 





1 


AF 

















1 


1 

















1 


1 



* AS = Australia. NG = New Guinea, SA = South America. AF = Africa. 




120 KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 

3. Solve the area relationships. To accomplish this, we must have an equivalent of 
character polarity decisions. This equivalent is the "zero vector ancestor" for each group. 
Specifically, we consider all "1" values as analogous to apomoiphies by inspecting the 
outgroup and seeing that it does not occur in the area or by assuming that any distributional 
data we might gather that relate to events prior to the origin of the ancestor of each group is 
irrelevant. Just as in character analysis, the confidence we have in our hypothesis of area 
relationships depends on the confidence we have in our polarity assignments. The solution 
to this matrix is shown in Fig. 7.5. Tliis is the history of vicariance based on phylogenetic 
and biogeographic data. 

AS NG SA AF 



AS + NG 

AS + NG + SA 

AS + NG + SA + AF 



Fig. 7.5. — The relationships among four continents as shown by the lizard and frog phylogenetic and 
biogeographic data, with "ancestral areas" labeled at each node. 



Missing Taxa 

Not all clades are distributed over all areas. There are several reasons why a representative 
of a group may not be in an area. 1) The group might be absent because the ancestor of the 
group never inhabited the area. 2) The group might be absent because a member of the group 
went extinct in the area. 3) The group might really be present but was not sampled. The 
correct explanation for the absence of a group from an area will vary from group to group. 
Several strategies might be employed to deal with the problem. 

Strategy 1. — Throw out the group. This strategy might be employed if one suspects that 
sampling effort is so low that the actual distributions of members of the group are not well 
enough known to employ them in an analysis. 

Strategy 2. — Use the group. This strategy can be used if the investigator has reason to 
believe that the distribution data she has are robust enough to be employed. Sampling has 
been thorough enough so that the distribution of the group is fairly well to very well 
known. Any absence due to sampling error is probably randomized over all groups. 
Hypotheses of absence due to extinction or ancestral distribuUon (i.e., ancestor was not 
distributed over all areas included in the analysis) are viable explanations. 



COEVOLUTIONARY STUDIES 121 

Note the difference in the two strategies. If you employ the group but then conclude that 
the anomalies in its distribution relative to the general hypothesis you generate were due to 
sampling error, it would have been wiser not to use the group in the first place. Unfortu- 
nately, there are no hard and fast mles for employing either strategy. Strategy 2 is the best to 
use when there is any question about actual distributions. The only effect of introducing such 
a group, if the distribution of its members are at variance with the general hypothesis 
generated, should be an increase in tree length. 

hi Example 7.3, we will assume that you are satisfied that strategy 2 has been employed. 
Your assumption is that although absences may be due to sampling error, you have made 
some effort to discover members of groups that are missing in particular areas. 

Example 7.3, — Missing taxa. 

Taking the same areas as in Example 7.2, we now consider the distributional data 
provided by two additional groups. Bird (B) and Worm (W). 

Bl: Australia (AS) 

B2: New Guinea (NG) 

B4: Africa (AF) 

W1:AS 

W3: South America (SA) 

W4:AF 

Note that each group is missing a representative in one area. The relationships of these 
species are shown in Fig. 7.6. We proceed as follows. 

1. Assign ancestors to the tree (Fig. 7.6). 

2. Prepare the matrix (Table 7.4). The cells in the matrix that have dashes (-) are cells for 
which there are missing data. We have to decide how to code these missing data. Recall the 
earlier discussion about observational and inferential data scores. Turning to the Bx/SAcell 
in the matrix, if we decide to code the Bx/SA cell as "0," we are making the explicit 
inference that the ancestor of Bird was not present in South America. If we code it as " 1 ," we 



B1 B2 B4 





Fig. 7.6. — ^Therelationshipsof agroupof (a)birds (B) and (b) wonms (W). with ancestral nodes labeled 
(x, y, z) (Example 7.3). 



122 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



Table 7.4. — Data matrix for birds (B 1 , B2, 34) and wonns ( W 1 , W3. W4) and their ancestral taxa (x, y, z) 
(Example 7.3). 















Taxa 












Area* 


Bl 


B2 


Bx 


B4 


Bz 




Wl 


W3 


Wy 


W4 


Wz 


AS 


1 





1 





1 




1 





1 





1 


NG 





1 


1 





1 










- 





- 


SA 








- 





- 







1 


1 





1 


AF 











1 


1 













1 


1 



* AS = Australia, NG = New Guinea. SA = South America. AF = Africa. 

are making the explicit inference that there is a member of Bird in South America or, 
alternatively, that there was a member of Bird there and it has gone extinct (i.e., we assign 
"1" as a phantom code indicating that we believe there is a missing ancestor and descen- 
dant). None of these alternatives are based on data nor on assumptions about the additive 
nature of descendant ranges relative to ancestral ranges. All are reasonable altemative 
hypotheses, at least at this stage in our analysis. We can remain neutral to these possibilities 
by assigning a missing data code ("?") to this cell. We proceed to each cell in the matrix and 
assign the missing data code to each ancestral column when a member of that group is not 
present in the area (Table 7.5). 

3. We then solve for the area relationships. When doing this we must take into account 
the "missing data" cells and their altemative inteipretations under different kinds of optimi- 
zation. Figure 7.7 shows three equally parsimonious interpretations using DELTRAN. 
Figure 7.7a shows the effect of delaying the "appearance" of Bx. If this is our choice, then 
we cannot delay the appearance of Wy. Figure 7.7b shows the effect of delaying the 
"appearance" of Wy. If we delay Wy, we cannot delay Bx. Finally, Fig. 7.7c summarizes the 
conflict between the distributions of both characters. Note that in neither case is there direct 
evidence for the grouping AS+NG-f-SA. Rather, the grouping is a by-product of the 
interactions between character distributions. Also note that even using DELTRAN neither 
Bz nor Wz is effectively delayed. Tliis is another interaction effect. (If you try ACCTRAN, 
you will find that both Bx and Wy are successfully accelerated but that the effect is to 
provide no character support for the groupings AS-i-NG or AS-i-SA.) 

Table 7.5. — Data matrix from Table 7.4 moditied to show neutrality about missing data. 















Taxa 












Area 


Bl 


B2 


Bx 


B4 


Bz 




Wl 


W3 


Wy 


W4 


Wz 


AS 


1 





1 





1 




1 





1 





1 


NG 





1 


1 





1 










9 





9 


SA 








9 





9 







1 


1 





1 


AF 











1 


1 













1 


1 



COEVOLUTIONARY STUDIES 



123 



^ ^ q,^ ^ 



^ C,^ ^ ^ 



^ # oT ^ 



— Bx 



— Wy 



Wy 



Bx 



--Bz 
-- Wz 



Bx 
Wy 



Bz 
+ Wz 



TBz 
Wz 



Fig. 7.7. — Three alternative trees (a-c) derived from analyzing the bird and worm data. Note that the 
synapomoiphies supporting AS+NG (tree a) and AS+S A (tree b) are illusions resulting from "missing data" 
cells in the matrix. 



EXERCISE 7.3.— Some moths. 

Examine the distiibution data listed below and the phylogeny shown in Fig. 7.8. 

1 . Prepare a data matrix for this group using the original four areas (AS, NG, SA, AF). 

M1:AS 
M2:NG 
M3: SA 

2. Add these data to the data concerning Bird and Womi (Table 7.5) and solve the area 
relationship problem. 



M1 



M2 



M3 




Fig. 7.8. — The relationships among three moths (M) with ancestral nodes labeled (x, y) (Exercise 7.3). 



124 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



Widespread Species 

A widespread species is one that is found in two or more areas. Obviously, this designa- 
tion is relative to the distributions of species in other groups. A group with a widespread 
species shows less endemism than a group with different species in all the areas analyzed. A 
species might be widespread for several reasons. 1) The species did not respond to 
geographic subdivision by speciating (failure to speciate). 2) The species dispersed into one 
of the areas (dispersal). 3) The species is really two species, but the investigator has not 
detected this fact (identification error). 

Given that you have analyzed the phylogenetic relationships among members of the 
group carefully and have done your best to identify each species, the only coding strategy 
you can employ is to code each species where it occurs. Example 7.4 demonstrates how to 
code for such cases. 

Example 7,4. — Coding widespread taxa. 

Continuing our example from Humphries and Parenti (1986), we consider two groups 
having widespread species: Tree (T) and Fish (Fi). The phylogenies of the two groups are 
shown in Fig. 7.9. The distribution of species in each group is listed below. 

Tl: Australia (AS) 

T2: New Guinea (NG) 

T34: South America (SA), Africa ( AF) 

Fil4:AS,AF 

Fi2: NG 

Fi3: SA 

We proceed as follows. 

1 . Assign ancestors to the trees (Fig. 7.9). 

2. Prepare the data matrix (Table 7.6). In this case, we will not assume that widespread 
species are either useful or not useful for solving the area relationships. Therefore, both the 
area and the ancestor will be assigned " I " when a descendant is present. 



T1 



T2 



T34 



Fi14 



Fi2 




Fi3 




Fig. 7.9. — The relationships within a group of (a) trees (T) and (b) fishes (Fi), with ancestral nodes labeled 
(x. y) (Example 7.4). 



COEVOLUTIONARY STUDIES 



125 



Table 7.6. — Data matrix for trees (Tl, T2, T34) and fish (Fil4, Fi2, Fi3) and their ancestral taxa (x, y) 
(Example 7.4). 















Taxa 












Area* 


Tl 


T2 


Tx 


T34 


Ty 




Fil4 


Fi2 


Fix 


Fi3 


Fiy 


AS 


1 





1 





1 




1 





1 





1 


NG 





1 


1 





1 







1 


1 





1 


SA 











1 


1 













1 


1 


AF 











1 


1 




1 





1 





1 



* AS = Australia. NG = New Guinea, SA = South America, AF = Africa. 

3. Solve the area relationships. In this case, there are three possible trees, each with 15 
steps and a CI of 0.833. Rather than present one or more of these trees, we are going to map 
the distributions of members of Tree and Fish onto the area relationships shown by Lizard 
and Frog (Example 7.2). This result is shown in Fig. 7.10. (Considering only the Tree and 
Fish data, this tree is 13 steps and has a CI of 0.769.) 

The tree in Fig. 7.10 reveals some interesting problems. Let us consider the distributions 
of T34, Fix, and Fil4 in the light of the three reasons for widespread species listed above. 
Reasons 1 and 3 lead you to expect that the species can be mapped as monophyletic on the 
area cladogram. In neither case, however, will the outcome affect the area cladogram. 
Reasons 1 and 2 might be expected if the species in question is paraphyletic when mapped 
on the area cladogram. Reason 1 is a little scary because acceptance of the hypothesis leads 
to accepting the species as an ancestor. Reason 2 would be expected if the species is 
polyphyletic when mapped on the area cladogram. We can be fairly confident that Fish 14 
has dispersed from Australia to Africa, but we are not very confident that Tree 34 dispersed 
or simply failed to respond to the vicariance event that separated South America and Africa. 



^* 



# 



--T1 --T2 



Fi14 



Fi2 



Tx 
Fix 



+ Fi3 
134 



^ 



^ T34 
Fi14 
4 Fix 



Ty 

+ Fiy 

Fig, 7,10, — A hypothesis of the relationships among four areas with the occurrence of trees and fishes 
mapped on the hypothesis. Tick marks denote unique distributions, black dots denote homoplasy. 



126 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

EXERCISE 7.4.— Ferns. 

Examine the distributional data and phylogeny (Fig. 7.11) for the group Fern (Fe). 
Prepare a data matrix for this group. Combine this with the data matrix for Tree and Fish and 
solve the area relationships using DELTRAN. Tip: There are three DELTRAN trees and 
three ACCTRAN trees, but solve only for DELTRAN. 

Fel2: Australia (AS), New Guinea (NG) 
Fe3: South America (SA) 
Fe4: Africa (AF) 



Fe12 Fe3 Fe4 




Fig. 7.11. — The relationships among three species of ferns (Fe), with ancestral nodes labeled (x, y) 

(Exercise 7.4). 



EXERCISE 7.5.— Combining the matrix. 

Combine the matrices for all of the groups inhabiting the four continental regions we have 
been working with (a total of eight groups). Solve the area relationships using DELTRAN. 
Examine the distribution of each group on the independent vaiiable (the cladogram of areas) 
and see if you can detect cases where a strict vicariance interpretation should not be 
followed. 

Sympatry within a Clade 

Sympatry within a clade occurs when two members of the clade inhabit the same area. 
How to treat sympatry between members of the same clade is still an open issue, and we 
have not designed exercises for this aspect of vicariance biogeography. Three suggestions 
have been made. Wiley (1988a) simply coded sympatric members of a clade as being 
present in the area. Kluge (1988b) suggested that in such cases one of the distributional 
pattems is younger than the origin of the areas. Thus, one of the members of the clade cannot 
furnish corroboration for the area hypothesis. This is certainly true, but the problem is, which 
member? Brooks (1990) has suggested another strategy, assigning two codes to the area 
where two members of the clade are sympatric. So, if Aus aus andAus bus both inhabit Area 



COEVOLUTIONARY STUDIES 127 

1 , split Area 1 into two areas { 1 and 1 '). Then find out wiiere 1 ' falls out in the analysis. How 
the two species are related to each other and how the two areas are related to each other can 
then be addressed. 

The Analogy between Phylogenetics and Historical Biogeography 

You can see that the computational similarities among phylogenetic systematics, 
vicariance biogeography. and studies of coevolution can be very impressive. However, the 
biological basis behind each area of study is different. For example, the relationship between 
characters and taxa is much more direct than the relationship between areas and taxa. Look 
at Fig. 7.10 again. Note that one of the ancestors in the Fish group is actually plotted as 
polyphyletic ( ! ). In character analysis we would simply conclude that the distribution was 
achieved by convergent evolution. This is allowed. However, the origin of a single species 
twice is not allowed. Several authors have called attention to these problems, including 
Wiley (1987c), Cracraft (1988), Sober (1988b), and Brooks (1990). 

Chapter Notes and References 

1. The use of a matrix to describe the shape of a tree was independently proposed by 
Farris (1969), Phipps ( 1971 ), and Williams and Clifford ( 1971 ); see Farris (1973). 

2. We have used additive binary coding throughout this chapter. This is not necessary, 
hideed, it is not necessary to enter the terminal taxa into a matrix because these data do not 
affect tree topology (cf. Kluge, 1988b; Mayden, 1988b). Further, we suggest that when you 
use a suitable computer program, you experiment with inputting your transfonnation series 
in such a manner that evolution is irreversible. See Mayden (1988b) for a discussion of 
possible problems in coding characters. 

3. Brooks ( 198 1 ) first proposed a technical solution to Hennig's parasitological method, 
and this led directly to the vicariance biogeographic and coevolutionary techniques pre- 
sented here. 

4. Biogeographic and coevolutionaiy techniques are virtually identical, as shown by 
Brooks (1985, 1988) and Wiley (1987d). A considerable body of literature on vicariance 
biogeographic methods and applications is summarized in Wiley (1988a). Parsimony 
methods are discussed in detail by Wiley (1987c, 1988a,b), Zandee and Roos (1987), and 
Kluge (1988b). Application of vicariance methods to the study of speciation include Wiley 
(1980, 1981a), Cracraft (1983, 1986). Mayden (1985, 1988b), Wiley and Mayden (1985), 
Brooks and McLennan (1991), and Siegel-Causey (1991). 

5. See McLennan et al. ( 1 988) for an example of the use of behavioral characters. For a 
review of a more general use of these methods in coevolutionary studies, see Brooks and 
McLennan (1991). 



128 KU MUSEUM OF NATURAL HISTORY. SPECIAL PUBLICATION No. 19 

Quick Quiz Answers 
Biogeography 

1. (2+3)+( 1 ). This is a short way to note the relationships. 

2. Presence/absence data use only the first three columns of the data matrix (Table 7. 1 ). When you use 
only these data, you see that the area relationships are not resolved. So the tree would be 1+2+3. Not very 
helpful, is it? 



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3:305-332. 



136 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



ANSWERS TO EXERCISES 



Two conventions are used in the exercises. For complicated trees, the topology is presented in parentheti- 
cal notation, and the synapomorphies are listed by group and the autapomorphies by tenninal taxon. In a few 
cases, a tree is so complicated that we present the tree with labeled intemodes. Characters supporting 
groupings are then shown by internode. IG refers to the ingroup. Characters for the IG refer to 
synapomorphies for an entire ingroup. 



CHAPTER 2 



EXERCISE 2.1.— Analysis oi'Siis. 

Each specific epithet has been abbreviated to its first letter. 



1. 



OG s t u V w 




((s, t. u. V. w. 1-1 )0G) 



OG s u t V w 




2-1 



((f, V. w.2-\)s. u, OG) 



OG V w s t u 




{(s, t, u:3-1.4-1)OG, V. w) 



OG u w V s t 




((s, /:5-1,6-1)OG, u. V, w) 



OG t u s V w 




7-1 



{(s. V. w7-1)0G, t, u) 



OG St u V w 




{(V, iv:8-1)OG,s, /, u) 



OG t u V w s 



9-1 




((s:9-1)0G, t. u.v.w) 



OG s u V w t 



10-1 




((f 10-1)OG, s, u, V. w) 



OG St u w V 



11-1 




((v. 11-1)0G, s. t.u.w) 



OG s t u V w 



12-1 




((w; 12-1)0G. s. t. u. V) 



138 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

2. Topology: {OG{vus . wiis)(uus(sus, tus))). Characters: (IG): l-l;(v«5, vv7«): 2-1; 7-1,8-1; (uus.susjus): 
3-\A-\-Asiis.tus):5-\,6-\:{siis):l-\,9-\:(tiis):2-\.lO-\:{vi(s): 11-1;(m7«): 12-1. © = character showing 
convergence/parallelism or reversal (homoplasies). 

OG V w u St 




3. Tree statistics: CI (tree) = 0.857; length = 14. CI (characters): 1, 3-6, 8-12 = 1.0; 2. 7 = 0.5. 



EXERCISE 2.2.— Analysis of Midae. 

Each generic name has been abbreviated to its first letter. 

1. Hypotheses of ,synapomorphy:(M,/V,0.e./?): \-\AlC):2-\AP.Q.R):3-\AM.P.Q.R):4-\:{M,N. 
O): 5-1, 10-1, 11-1: {M,N. O. Q): 6-1; (N. O. Q. R): 7-1: [P): 8-1; (A/): 9-1. 

2. Most parsimonious tree topology: (OG{P{R{Q{M(N . O)))))). Characters: (IG): 2-1, 3-1,4-1; (R, Q. M, 
N. 0): 1-1, 7-1: (Q. M.N. O): 6-1; (M. N. O): 3-0, 5-1. 10-1, 11-1; (A', O): 4-0: (P): 8-1; (M): 7-0, 9-1. 

3. Tree statistics: CI (tree) = "/u = 0.786; length = 14. CI (characters): 1, 2, 5, 6, 8-1 1 = 1.0; 3, 4, 7 = 0.5. 



EXERCISE 2.3.— Analysis of Ait.s. 

Each specific epithet has been abbreviated to its first letter. 

1, 2. There are two tree topologies and a total of four trees (two trees for each topology). All four trees 
share the following autapomorphies: (a): 6-1; (c): 10-1; (c/): 8-1; (e): 9-1. Note that character 7-1 is 
homoplasious for both taxa h and e on all four trees. 

Tree topology 1: {OG((a. h)c)(d. e)). Synapomorphies for tree lA — (a. h. c. d. e): 1-1, 2-1; {a. h. c): 3-1; 
(a. by. 2-0, 5-1. Synapomorphies for tree IB— (a. h. c. d. e): 1-1; {a. b. c): 3-1; (a, b): 5-1; (d. e): 2-1,4-1. 
Autapomorphies unique for tree IB — (c): 2-1. CI (characters): 1, 3-6, 8-10 = 1.0; 2, 7 = 0.5. 

Tree topology 2: (OG(a, b)(c{d, e))). Synapomorphies for tree 2A — (a. b. c. d. e): 1-1, 3-1; {a. b): 5-1; 
(c.d.e):2A\ id. c): 3-0, 4- 1 . Synapomorphies for tree 2B— (o, b. c. d. e): lAAa. b): 3-1. SAAc d. e): 2-1; 
{d. e): 4-1. Autapomorphies unique for tree 2B — (r): 3-1. CI (characters): 1, 2, 4-6, 8-10 = 1.0; 3, 7 = 0.5. 

3. Tree statistics: CI (tree) = 0.833; length = 12. 



ANSWERS TO EXERCISES 



139 



CHAPTER 3 



EXERCISE 3.1 



Did you label node 1? If so, you're wrong. Node 1 is the root and you cannot label it without using its 
outgroup. 









Nodes 






TS 


2 


3 


4 


5 


6 


OG 


3 
4 


a.b 
a 


b 
a 


b 
b 


a,b 

a.b 


b 
a 


decisive 
decisive 



M N O P Q R IG 




EXERCISE 3.2 



EXERCISE 3.3 



Nodes 



TS 


T 


U 


V 


W 


OG 


1 


a 


a.b 


b 


b 


decisive 


2 


a 


a 


a,b 


b 


decisive 


3 


a 


a 


a 


a,b 


equivocal 


4 


a,b 


a 


a.b 


a 


decisive 


5 


a 


a.b 


b 


a.b 


equivocal 


6 


a,b 


a 


a,b 


b 


decisive 



Nodes 



TS 


M 





P 


Q 


OG 


1 


a 


a.b 


b 


b 


decisive 


2 


a,b 


b 


b 


b 


decisive 


3 


a,b 


a 


a,b 


a 


decisive 


4 


a 


a.b 


a 


a 


decisive 


5 


a 


a 


b 


a.b 


equivocal 


6 


a 


a,b 


a.b 


a,b 


equivocal 



EXERCISE 3.4 



Nodes 



TS 


M 


N 





P 


Q 


R 


OG 


1 


a 


a.b 


b 


a.b 


a 


a 


decisive 


2 


b 


b 


b 


b 


a.b 


a 


decisive 


3 


a.b 


a 


a.b 


a 


a 


a 


decisive 


4 


b 


b 


a.b 


b 


a.b 


a 


decisive 


5 


b 


b 


a.b 


a 


a 


a 


decisive 


6 


b 


b 


b 


a.b 


a 


a 


decisive 



140 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



EXERCISE 3.5 



Nodes 



TS 


P 


Q 


R 


S 


T 


U 


V 


W 


X 


Y 


OG 


1 


a 


b 


a.b 


a 


a 


b 


b 


a.b 


a 


a 


decisive 


2 


a,b 


a 


a 


a 


a 


b 


b 


b 


b 


a.b 


equivocal 


3 


a 


a 


a 


a.b 


a 


b 


a.b 


a 


a 


a 


decisive 


4 


a.b 


a.b 


b 


b 


b 


a.b 


b 


a.b 


b 


b 


decisive 


5 


a 


b 


b 


b 


a.b 


b 


b 


a.b 


a 


a 


decisive 


6 


a 


b 


b 


b 


a,b 


b 


b 


b 


a.b 


a,b 


equivocal 



EXERCISE 3.6 

1 . Linear coding and additive binary coding. 

2. Nonadditive and mixed are not applicable because the character tree is linear. 

3. Note that the plesiomorphic character "1" is not listed because its vector is all zeros. 



Taxon 



Linear 



Additive binary coding 



OG 
A 
B 
C 



4. Tree:(OG(A(B,C))). 



EXERCISE 3.7 



1 . Nonadditive binary coding. 

2. Linear coding and additive binary coding cannot be used because the tree represents a branching TS. 
Mixed coding would not save many columns because the tree is symmetrical. 

3. 

Nonadditive binary coding 



Taxon 


m 


n 





p 


q 


r 


s 


OG 























A 






















B 





















C 





















D 















1 





E 















1 





F 


















1 


S 





















T 









1 











U 









1 











V 






1 














W 






1 















4. Tree: (OG(A(B, C, F(D, E))(S(T, U)( V, W)))). 



ANSWERS TO EXERCISES 



141 



EXERCISE 3.8 

1 . Nonadditive binary and mixed coding; mixed coding is preferred because it would save columns. 

2. Linear and additive coding will not work because the tree represents a branching TS. 
3. 









Nonadditive binary coding 








Mixed coding 




Taxon 


n 





P 


q 


r 


s 


t 


u 


n, q, s, t 


r 


u 


o,p 


OG 






































A 

























1 











B 




1 




















1 








1 


C 




1 


1 

















1 








-> 


D 




1 


1 

















1 








2 


M 
























2 











N 












1 











2 


1 








O 























3 











P 




















1 


3 





1 





Q 

















1 





4 











R 

















1 





4 












4. Tree: (OG(A(B(C. D))(M, N(0, P(Q, R))))). 



CHAPTER 4 



EXERCISE 4.1.— Bremer's Leysera data. 
1. 



Characters* 



Taxon 


1 


2 


3 


4 


5 


6 


OG 




















L. longipes 




















L. leyseroides 


1 


1 


1 


1 


1 


1 


L. tennella 


1 


1 


1 


1 


1 


1 


L. gnaphalodes 


1 


1 


1 


1 









* 1 = receptacle, 2 = floret tubules, 3 = pappus type, 4 = achene surface, 5 = pappus scales, 6 = life cycle. 



2. Specific epithets are abbreviated. Tree: (OG(lo(gn(ten, />')))). Characters within Lysera: {ly, ten, gn): 1-1 , 
2-1,3-1,4-1: {ten. ly): 5-]. 6-1. Synapomorphies of Lvicra: 2N = 8, solitary capitula on long peduncle. 

EXERCISE 4.2.— Siegel-Causey's cliff shags. 

1. OG vector: 1-0. 2-0/1. 22-0. 36-0/1. 39-0. 40-1. 42-0, 48-0, 63-0, 69-0, 78-0, 79-1. 81-0. 94-0, 97-0, 
100-1, 102-0/1, 110-0, 111-0, 112-0, 114-0, 120-0, 124-0/1, 131-0, 134-0. 

2. We did not consider TS 2. TS 36. TS 102, and TS 124 because their character decisions were equivocal 
at the OG node. 

Some specific epithets are abbreviated. Tree: {OGiipelag. uhle)(arisi(gaim(piinct.feath))))). 

Characters: (IG): 97-\:(pelag. wile): 134-1; (arist. gaim.pimct.feath): 63-1, 110-1; (gaim.pimct.feath): 
39-1. 40-0, 79-0, 94-1, 111-1, 112-1; (punct . feath): 1-1, 78-1; (pelag): 42-1; (wile): 69-1, 120-1, 131-1; 
(arist): 22-1. 100-0; (gaim): 48-1. 1 14-1: (pumt): none; (J'eath): 81-1. 



1 42 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 1 9 

3. TS 2 and TS 102 remain equivocal. Character 36-0 is a synapomorphy shared by i^aim. piinct, and 
feath. Character 124-0 is either a homoplasy shared by gaim and feath or a synapomorphy below gaii)i that 
reverses in piimt. 

EXERCISE 4.3 

D(A,B) = IIX(A./)-X(B,/)l 

:.|0-OI-Hll-ll-HlO-ll-Hll -01-^11-01 

= 3 
D(M.B) = 3 

EXERCISE 4.4 

1. D(M,A) = IIX(M,/)-X(A,/)l 

= 1 1 _ 01 -H 10 - 1 1 4- 10 - 01 -I- 10 - 01 + 10 - 01 
= 2 
D(M,B) = 3 

ANC ABC 

ANC 

A 2 - 

B 3 3- 

C 4 4 3- 

2. INT(A) = D[A.ANC(A)] 

= 2 

EXERCISE 4.5. — Bremer's Leysera data. 
1. Construct a data matrix. 

Transformation series 



Taxon 


1 


2 


3 


4 


5 


6 


Ancestor (ANC) 




















longipes (In) 




















leyseroides (/>') 


1 


1 


1 


1 


1 


1 


tennella (le) 


1 


1 


1 


1 


1 


1 


gnaphoides (gn) 


1 


1 


I 


1 









2. Calculate initial differences. 

D(ANC./h) = 

D(ANC,/y) = 6 

D(ANC.re) = 6 

D(ANC,g/0-4 



ANSWERS TO EXERCISES 



143 



3. Add the taxon (//;) with least difference to ancestor (ANC). 
/n(OOOOOO) 



ANC (0 00000) 



4. Add gnaphaloides. 



Transformation series 



Taxon 


1 


2 


3 


4 


5 


6 


ANC 




















gn 


1 


1 


1 


1 








In 




















A 





















5. Add either /v or te — let's use /v. 



D(ly.ln) = 6 

D(/v,i?n) = 2 

D(/yA) = 6 

D(/yANC) = 6 

D(/.v.//2) + D(/vA)-D(/rtA) 



D(/yJNT(/;i)) : 



DC/yJNTC?;!)) ■- 



D(/yJNT(A)) = 



6 + 6-0 



D(ly.gn) + D(/vA) - DignA) 



2 + 6-4 



D(/vA) + D(/vANC) - D(A.ANC) 



6 + 6-0 



gn(^ 11100) In 



A (0 0) 



ANC 



= 6 



Conclusion: add /y to INT (gn) 



144 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



6. Calculate a new ancestor and an intermediate tree. 
Transformation series 



Taxon 


1 


2 


3 


4 


5 


6 


A 




















gn 


1 


1 


1 


1 








ly 


1 


1 


1 


1 


1 


1 


B 


I 


1 


1 


1 









gn /y(1 1 1 1 1 1) 



B (1 1 1 1 0) In 



ANC 



7. Add tennella in tlie same manner. 

D(fe,/v) = D(//i,A) = 

V)(te.gn) = 2 D(B,A) = 4 

D(teS.) = 2 D(A.ANC) = 

D{te.ln) = 6 D((v,B) = 2 

D(te.A) = 6 D(gn.B) = 
D(re,ANC) = 6 

8. Calculate a new ancestor and find the final tree. 



Transformation series 



Taxon 



B 


1 


te 


1 


ly 


I 


c 


1 



gn 



D[/£',INT(x«)l = 2 
D[?e.INT(/v)] = 
D[ft',INT(B)]-2 
D[re.INT(/H)] = 6 
D[re,lNT(A)] = 6 



fe(l 11111) ly 



C (1 1 1 1 1 1) 



ANC 



EXERCISE 4.6.— Siegel-Causey's cliff shags. 

1. Calculate differences. 

D(OG,pelag) = 4 
D(0G,«n7e) = 6 
D{.OG,arist) = 6 

Begin with pelag. 
pelag 



D{OG, gaim) = 14 
D(OG, piiiict)= 12 

DiOG feaih) = 14 



OG 



ANSWERS TO EXERCISES 



145 



2. The next shortest are urile or arist. We picked arist (but it won't make any difference at the end). 
pelag arist 




Transformation series 



Taxon 1 2 22 36 39 40 42 48 63 69 78 79 81 94 97 100102 110 111112 114120124131134 



OG 








1 


1 


1 














1 











pelag 





1 


1 


1 


1 1 














1 








1 


arist 





1 


1 


1 


1 





1 








1 








1 


A 








1 


] 


I 














1 








1 



1000000100 
10 10 1 
1 10 10 
1000000100 



3. Now pick MnVe. 

D[ttn7e,INT(A)] = 5 

Diiirile.A) = 5 

D(urile.OG) = 6 

D[urileMT{pelag)] = 3 

Join urile to pelag by ancestor B. Calculate B. 

pelag urile 



arist 




D(urile,pelag) - 4 
D[unle.]KT[arist)] = 5 
D(unle. arist) = 10 



Transformation series 



Taxon 1 2 22 36 39 40 42 48 63 69 78 79 81 94 97 100102 110 111112 114 120124131134 



A 1 

pelag 10 1 

urile 10 1 

B 10 1 



01000001001 1 

01 100001001 1 

01000101001 1 

01000001001 1 



000000100 
10 1 
1 1 1 1 
10 1 



146 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



4. Pick punct to add next. 

Dipunct. OG)^ 12 
D[punct.lNT(A)] = 1 1 
D(pimct.B)= 13 
Dlpunct.lNTipelag)] = 13* 



D[punct,INT(arist)] = 9 
D(piinct.A) = 1 1 
D[punct.lNT(wile}]= 13* 
D{punct,pelag) = 14* 



D(punct,arist) = 12 
D|/7(wrt.INT(B)] = 11 
D(pimct.urile) = 16* 



(*As per instructions, it is not necessary to calculate these distances because the distance oi punct to IKV(arist) is 
less than the distance between p/(rtc7 and INT(B). Therefore, there is no need to calculate distances between pwwf/ and 
terminal taxa above B.) 



Add punct to arist by ancestor C. Calculate C. 



pelag 



urile 



punct 



arist 




Transformation series 



Taxon 1 2 22 36 39 40 42 48 63 69 78 79 81 94 97 100102 110 111112 114120124 131134 



A 00010100000100 

pimcl 10001000101001 

arist 001 10100100100 

C 00010100100100 



1 1000000100 
110 1110 10 
10 1 10 10 
1 10 10 10 



5. Pick gaim to add next. 

Digaim.OG) = 14 

D[ga/w.INT(B)] = 13 

Digaim.C)= 11 

D[gaim.lNT{arist)] = 10 



D[ga/w.INT(A)] =: 13 
D(gaim.B)= 15 
Dlgaim.YNJipunct)] - 4 
D(gaim, arist) = 12 



Add gaim to pwu-t through D. Calculate D. 

purict gaim 



pelag 




arist 



D(gairn.A}= 13 
D[ga/w,INT(C)] = 11 
T>{gaim, punct) - 6 



OG 



ANSWERS TO EXERCISES 



147 



Transformation series 



Taxon 1 2 22 36 39 40 42 48 63 69 78 79 81 94 97 100102 110 111112 114120124131134 



C 








1 


1 








1 





1 








1 


1 


punct 1 











1 








I 


1 








1 


1 


1 


gaim 











1 





1 


1 











1 


1 


1 1 


D 











1 








[ 











1 


1 


1 



10 10 

1 1 10 10 

11110 

1 1 10 10 



6. Finally ( ! ) add feath to the tree. 

Dtfeath.OG) = 14 

D\feath.lNT{B}] ^ 13 

D(feath.C)= 11 

D\feath.mT(punct)] = 2 

Difeath.gaim) = 6 

Add feath to punct through E. Calculate E. 

punci feath 



D[/eaf/;,INT(A)] = 13 
D{feathB)= 15 
D[/eaf/!.INT(D)] = 4 
Y>{feath, punct) = 2 



pelag 




arist 



Transformation series 



D(feath.A)= 13 
D[feath.lNT{C)]= 11 
D(feath,D) = 4 
D\feath.]NT{gaim)] = 3 



Taxon 1 2 22 36 39 40 42 48 63 69 78 79 81 94 97 100102 110 111112 114 120124131134 



D 00001000 

puncl 10 10 

feath 1 ,1 

E 10 10 









1 


1 1 I 


1 


1 





1 








1 


1 


110 1 


1 


1 





1 








1 


1 1 


110 1 


1 


1 














1 


1 ; 


110 1 


1 


1 





1 






EXERCISE 4.7 

1 10 





148 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



EXERCISE 4.8 



110 10 1 




EXERCISE 4.9 




EXERCISE 4.10 



DELTRAN Optimization 




X 



M 



N 1 



C 0.1 



Step 1 : Label ancestor nodes and 
assign root state. (This is a trivial 
tasl(.) 



A O 1 C 0.1 



B 0,1 

Step 2: Reroot at B, the next ancestor 
up from the root, and recalculate the 
MPR. MPR(B) = (0,1), 



B 0.1 MO N 1 



C 0,1 

Step 3: Reroot at C. the last ancestor 
in the tree; recalculate the MPR. 
MPR(C) = (0.1). 



10 1 
X O M N 






Step 4: Label all ancestor nodes with 
appropnate MPRs. 



Step 5; Assign ancestor states on 
upward pass. 



Step 6: Place character changes on 
appropriate branches following the 
state of the ancestral node. 



ANSWERS TO EXERCISES 



149 



ACCTRAN Optimization 



10 1 

X O M N 




Step 1 : Assign ancestral stales on 
downward pass. 




Step 2: Assign states on upward 
pass. 




Step 3: Place character states on 
tree following states of ancestral 
nodes. 



EXERCISE 4.11 



DELTRAN Optimization 




MPR(A) = (0) 
MPR(B) = (0) 
MPR(C) = (0) 
IVlPR(D) = (0,1) 
tVlPR(E) = (0,1) 
MPR(F) = (0) 

Step 1 : Calculate MPRs for each ancestor node. 




Step 2: Label each ancestor node with appropriate 
MPR. 








110 



Step 3: Assign state of ancestor node by upward 
pass rules. 




Step 4: Label tree following the ancestor character 
states. 



150 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



ACCTRAN Optimization 



1 10 



110 





Step 1 : Assign ancestor states by downward pass 
rules. 



Step 2: Assign states of ancestor nodes by upward 
pass rules. 



1 10 




Step 3: Label tree following ancestor states. 



CHAPTERS 

EXERCISE 5.1 

Tree: (X(A(B, C))). M = 8, S = 8, G = 10, tree length = 8, CI = 1 .0, R = 1.0, RC = 1.0. 

EXERCISE 5.2 

M = 5, S = 6, G = 9, tree length = 6, CI = 0.833. R = 0.750. RC = 0.625. 

EXERCISE 5.3 

M = 8, S = 9. G = 1 1 , tree length = 9, CI = 0.889, R = 0.667, RC = 0.593. 

EXERCISE 5.4 

Tree: (OG(0(M, N))). M = 10, S = 12, G = 15. tree length = 12, CI = 0.833. R = 0.600, RC = 0.500. 

EXERCISE 5.5 

Tree: (OG(A((B, C)((D, E)F)))). M=ll,S=15,G = 26,tree length = 15. CI = 0.733. R = 0.733. RC = 0.538. 

EXERCISE 5.6 

Strict consensus tree: (A. B(C, D. E)). 

EXERCISE 5.7 

Adams consensus tree: (A{B, C. D)). 



ANSWERS TO EXERCISES 151 

EXERCISE 5.8 

Adams consensus tree: (A, B(C. D. E)). 

EXERCISE 5.9 

Majority consensus tree: ((W. V)(X(Y, Z))). 

EXERCISE 5.10 

Strict consensus tree: ((A, B. C)(D, E, F, G, H)). 
Adams consensus tree: ((A, B. C){D. E, H(F. G))). 
Majority consensus tree: ((A, B, C)(.D(E, F(G, Hj))). 



CHAPTER 6 

EXERCISE 6.1. — Phylogeny of the Recent tetrapod vertebrates. 

Classification a: (Lissamphibia. Aves, Mammalia(Chelonia. Lepidosauria. Crocodylia)). 45 trees pos- 
sible. Classification is logically inconsistent with the phylogeny. "Reptilia" includes Chelonia, Lepidosauria, 
and Crocodylia. 

Classification b: (Lissamphibia, Mammalia. Chelonia. Lepidosauria, Crocodylia, Aves). 945 trees pos- 
sible. Classification is logically consistent with the phylogeny. 

Classification c: Tree topology is identical to that of the tree in Fig. 6.10. One tree possible. Classification 
is logically consistent with the phylogeny. 

EXERCISE 6.2.— Phylogeny of the land plants. 

Tree: ((Anthoceropsida, Marchantiopsida, Bryopsida)(Psilotophytina, Lycopodophytina, Sphenophytina, 
Pteridophytina(Cycadopsida, Pinopsida, Ginkgopsida, Gnetopsida. Angiospermopsida))). Bryophyta in- 
cludes Anthoceropsida. Marchantiopsida. and Bryopsida. Tracheophyta includes Psilotophytina, 
Lycopodophytina. Sphenophytina. Pteridophytina, and Spermatophytina. Spermatophytina includes 
Cycadopsida. Pinopsida. Ginkgopsida, Gnetopsida, and Angiospermopsida. 33,075 trees possible. Classifi- 
cation is logically inconsistent with the phylogeny. 

EXERCISE 6.3 

We have used each letter on the tree to stand for a genus. It doesn't matter if you picked different categories 
or different endings to the names. Instead, look for conformity in the niimher of categories. It shouldn't 
matter for systematic relationships if you have different names or endings. 

1 . Sequence convention: 

Order A-S 
Family A-F 
Genus F 
Genus E 
Genus D 
Genus C 
Genus B 
Genus A 



152 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

Family M-S 
Genus M 
Genus A' 
Genus O 
Genus P 
Genus Q 
Genus R 
Genus S 

Hierarchical levels: 3. 

2. All branch points named: 

Order A-S 
Suborder A-F 
Family F 

Genus F 
Family A-E 
Subfamily E 

Genus E 
Subfamily A-D 
Tribe D 

Genus D 
Tribe A-C 
Subtribe C 
Genus C 
Subtribe AB 
Genus A 
Genus B 
Suborder M-S 
Family M 

Genus M 
Family N-S 
Subfamily N 

Genus A^ 
Subfamily O-S 
Tribe O 

Genus O 
Tribe P-S 
Subtribe P 
Genus P 
Subtribe Q-S 
Supergenus Q 

Genus Q 

Supergenus RS 

Genus R 

Genus S 

Hierarchical levels: 8. 

3. Five hierarchical levels are saved by using the sequencing convention and not naming all the branch 
points. 



ANSWERS TO EXERCISES 153 

EXERCISE 6.4 

We have used each letter on the tree to stand for a genus. It doesn't matter if you picked different categories 
or different endings to the names. Instead, look for conformity in the number of categories. 

Order A-U 

Genus A' (incertae sedis) 
Suborder M (sedis miitabilis) 

Genus M 
Suborder A-H [sedis miitabilis) 
Genus H (incertae sedis) 
Family G 

Genus G 
Family F 

Genus F 
Family A-E 

Subfamily C (sedis mulabilis) 

Genus C 
Subfamily D (sedis mutabilis) 

Genus D 
Subfamily E (sedis mutabilis) 

Genus E 

Subfamily AB 

Genus A 

Genus B 

Suborder P-U (sedis mutabilis) 

Family P 

Genus P 
Family Q 

Genus Q 
Family R 

Genus R 
Family S-U 

Genus 5 (sedis mutabilis) 
Genus T (sedis mutabilis) 
Genus U (sedis mutabilis) 

EXERCISE 6.5 

1. Order Tusiformes 

tGenus X (incertae sedis) 
Suborder A-F 
tFamily F 

Genus F 
Family A-E 
tSubfamily E 
Genus E 
Subfamily A-D 
Tribe D 
Genus D 



1 54 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 1 9 

Tribe A-C 

Genus A (sedis mutabilis) 
Genus B (sedis mutabilis) 
Genus C (sedis mutabilis) 
Suborder M-S 

Genus P (incertae sedis) 
Family M-O 
Tribe O 

Genus 
Tribe MN 
Genus M 
tGenus N 
Family Q-S 
Tribe Q 

Genus Q 
Tribe RS 
Genus R 
Genus 5 

2. Order Tusiformes 

tGenus X (incertae sedis) 

Suborder A-F 

Plesion F 

Plesion E 

Family D 

Genus D 
Family A-C 

Genus A (sedis mutabilis) 
Genus B (sedis mutabilis) 
Genus C (sedis mutabilis) 
Suborder M-S 

Genus P (incertae sedis) 
Family M-O 
Tribe O 

Genus O 
Plesion N 
Tribe M 
Genus M 
Family Q-S 
Tribe Q 

Genus Q 
Tribe R 

Genus R 
Tribes 
Genus S 

3. In #1, 6 categories and 29 entries are required in the classification: in #2, only 5 categories (a plesion 
doesn't count) and 26 entries. On a strictly numerical basis, use of Convention 7 is more justified than not 
using it. Use of plesions simplifies the classification without implying possibly unknown relationships. 



ANSWERS TO EXERCISES 



155 



EXERCISE 6.6 

1. 



/ .<-->' :,» ,/ .,/ ..i. 



/////// 



30-1 




2. 



N. harrisi I P. penicillatus P. africanus 

P. auritus P. pygmaeus 

P. olivaceus 
P. carbo 

Venn diagram of traditional classification 



N. harrisi P. penicillatus 1 



P. auritus 
P. olivaceus 



P. carbo 



P. africanus 
P. pygmaeus 



Venn diagram of phylogeny 



Q 



N. harrisi 



P. penicillatus 



D 



p. auritus 
P. olivaceus 



P. carbo 



P. africanus 
P. pygmaeus 



Venn diagram of consensus 



156 KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 

3. There are many classifications depending upon the nature of the categories you wish to use. Here is one, 
using the hsting convention. 

Firstgenus 

F. pygmaeus 

F. afiicanus 
Secondgenus 

S. penicillatus 

S. Imrrisi 
Thirdgenus 

T. carbo 

T. aurirus 

T. olivaceus 



CHAPTER 7 

EXERCISE 7.1. — Luxilus zonatus species group. 

1. Taxon tree: ((cardinalis, pilsbryi)zonatus). Ancestral taxa are labeled x and y. 

Taxa 



Area 


zonatus 


y 


cardinalis 


pilsbryi 


X 


1 





1 


1 





1 


2 





1 





1 


1 


3 


1 


1 












2. Areatree: ((1,2)3). 

EXERCISE 7.2. — Fundulus nottii species group. 

Each of the specific epithets used in this exercise have been abbreviated to their first three letters. W, X, Y, 
and Z are ancestral taxa. 

Taxa 



Area 


LIN 


ESC 


NOT 


w 


X 


BLA 


DIS 


Y 


z 


I 


1 











1 













2 





1 





1 


1 













3 








1 


1 


1 













4 

















1 





1 




5 




















1 


1 





Area tree: ((1(2,3))(4,5)). Trees are congruent. 



ANSWERS TO EXERCISES 157 

EXERCISE 7.3.— Some moths. 
1. 

Taxa 



Area 


Ml 


M2 


Mx 


M3 


My 


AS 


1 





1 





1 


NG 





1 


1 





1 


SA 











1 


1 


AF 








? 





? 



2. Onetree:(((AS,NG)SA)AF). 

ACCTRAN optimization. Synapomorphies— (AS, NG): Mx-1; (AS, NG, SA): Bx-1, Wy-l; (AS, NG, 
SA, AF): Bz-1. Wz-1, My-1. Autapomorphies— (AS): Bl-1. Wl-1, Ml-I: (NG): B2-1, M2-1: (SA): M3-1, 
W3-1;(AF):B4-1,W4-1. 

DELTRAN optimization. Synapomorphies— (AS, NG): Mx-1, Bx-1; (AS, NG, SA): My-1, Wy-l; (AS, 
NG, SA, AF): Bz- 1 . Wz- 1. Autapomorphies identical to those of ACCTRAN. 



EXERCISE 7.4.— Ferns. 
1. 

Taxa 



Area 


Fel2 


Fe3 


Fex 


Fe4 


Fey 


AS 


1 





1 





1 


NG 


1 





1 





1 


SA 





1 


1 





1 


AF 











1 


1 



2. DELTRAN optimization. Tree 1: (((AS, NG)SA)AF). Synapomorphies— (AS, NG): Fel2-1, Tx-1, 
Fix-1; (AS, NG, SA): Fex-1; (AS, NG, SA, AF): Ty-1, Fiy-1, Fey-1. Autapomorphies— (AS): Tl-1, Fil4-1; 
(NG): T2-1, Fi2-1; (SA): T34-1, Fi3-1, Fe3-1. 

Tree 2: (((AS, NG)AF)SA). Synapomorphies— (AS, NG): Fe 12-1, Tx-1, Fex-1; (AS, NG, AF): Fix-1; 
(AS, NG,AF,SA): Fey-1, Fiy-1, Ty-1. Autapomorphies— (AS): Fil4-1, Tl-1; (NG):Fi2-l,T2-l;(AF):T34-l, 
Fe4-l,Fil4-I;(SA):T34-l,Fi3-l,Fe3-l, Fex-1. 

Tree 3: ((AS, NG)(SA, AF)). Synapomorphies— (AS, NG): Tx-1, Fix-1, Fel2-1, Fex-1; (SA, AF): T34-1; 
(AS, NG,AF, SA): Ty-1, Fiy-1, Fey-1. Autapomorphies— (AS): Tl-1, Fil4-1; (NG):T2-l,Fi2-l; (SA): Fi3-1, 
Fe3-1, Fex-1; (AF): Fi 14-1. Fix-1, Fe4- 1. 

Tree statistics: CI = 0.833, length = 18, R = 0.500, RC = 0.417. 



158 



KU MUSEUM OF NATURAL HISTORY, SPECIAL PUBLICATION No. 19 



EXERCISE 7.5.— Combining the matrix. 

The black dots are homoplasies. The arrow suggests moving ancestral taxon Fiy-1 up one branch. If 
ancestor Fiy-1 moves up, then ancestor Fix-1 should be removed from Africa because by placing ancestor 
Fiy-1 higher it can no longer give rise to Fix-1 in Africa. The higher position of Fiy-1 also suggests that 
Fil4-1 dispersed to Africa. The T34-1 distribution may represent dispersal or persistence of a widespread 
species. If dispersal, the direction cannot be determined. 



kF 


SA 


NG 


AS 


■ F4-1 


. 


F3-1 




- F2-1 




■ L4-1 


- 


L3-1 




- L2-1 




■ B4-1 


• 


W3-1 




- B2-1 




W4-1 


- 


M3-1 




- M2-1 


. 


1 T34-1 


( 


» T34-1 




- T2-1 




Fe4-1 


- 


Fe3-1 




L Fi2-1 




1 Fi14-1 


- 


FI3-1 




( 


• 


1 Fix-1 
















• Fx-1 












■ Lx-1 










- Bx-1 










- Mx-1 










• Tx-1 










• Fe12-1 










• Fix-1 






-Fy-1 
-Ly-1 

- Wy-1 

- My-1 

■ Fex-1 










^ 










• Fz-1 








■ Lz-1 








■ Bz-1 








- Wz-1 






: 


> 


Ty-1 
Fey-1 








riy- 1 









F1-1 

L1-1 

B1-1 

W1-1 

M1-1 

T1-1 

Fi14-1 



Date Due 



j4QV-4=^.499^ jtitr?^^=^m 



3. 



7. 



Maintenar 
and Barr> 
ISBN: 0-8 

The Nature 
and James 
0-89338-0 

A Diapsid 
Reisz, pp. i- 




NtfiR;29199i 



MAY 3 1 2000 



les B. Murphy 
scember 1978. 



y L. Armstrong 
r 1979. ISBN: 



By Robert R. 
.J-011-3. 



9. The Ecological Impact of Man on the South Florida Herpetofauna. By 
Larry David Wilson and Louis Porras, pp. i-vi, 1-89, 8 August 1983. 
ISBN: 0-89338-018-0. 

10. Vertebrate Ecology and Systematics: A Tribute to Henry S. Fitch. 
Edited by Richard A. Seigel, Lawrence E. Hunt, James L. Knight, Luis 
Malaret, and Nancy Zuschlag, pp. i-viii, 1-278, 21 June 1984. ISBN: 
0-89338-019-0. 

13. Geographic Variation among Brown and Grizzly Bears {Ursus arctos) 
in North America. By E. Raymond Hall, pp. i-ii, 1-16, 10 August 
1984. 

15. Spring Geese and Other Poems. By Denise Low, pp. 1-84, September 
1984. ISBN: 0-89338-024-5. 

18. A Checklist of the Vertebrate Animals of Kansas. By George D. 
Potts and Joseph T. Collins, pp. i-vi, 1^2, September 1991. 
ISBN: 0-89338-038-5. 



^4833 



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OH83 .C73 1991 

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Harvard MCZ Library 



3 2044 062 409 495